MEI Core 1. Basic Algebra. Section 1: Basic algebraic manipulation and solving simple equations. Manipulating algebraic expressions

Transcription

1 MEI Core Basic Algebra Section : Basic algebraic manipulation and solving simple equations Notes and Examples These notes contain subsections on Manipulating algebraic expressions Collecting like terms Expanding brackets Factorising Multiplication Algebraic fractions Linear equations Changing the subject of a formula Manipulating algebraic expressions Throughout your mathematics you will need to be able to manipulate algebraic expressions confidently. The examples below remind you of the important techniques of collecting like terms, removing brackets, factorising, multiplying, and adding, subtracting and simplifying algebraic fractions. Example Simplify the expression a + b a + b ab + a Collecting like terms There are three different types of like term in this expression. There are terms in a, terms in b, and a term in ab. Notice that the term in ab cannot be combined with either the terms in a or the terms in b, but remains as a term on its own. a + b a + b ab + a = a a + a + b + b ab = 4a + 5b ab In the example the expression has been rewritten with each set of like terms grouped together, before simplifying by adding / subtracting the like terms. You may well not need to write down this intermediate stage. For practice in examples like this one, try the interactive questions Collecting terms. MEI, 4/05/0 /8

2 MEI C Algebra Section Notes and Examples Expanding brackets Example Simplify the expressions (i) (p q) + (p + q) (ii) x(x + y) y(x 5y) Each term in the bracket must be multiplied by the number or expression outside the bracket. (i) (p q) + (p + q) = p 6q + 6p + q = 9p 4q (ii) x(x + y) y(x 5y) = x² + 6xy xy + 5y² = x² + 4xy + 5y² Multiplying out two brackets of the form (ax + b)(cx + d) gives a quadratic function.make sure that you are confident in this. Example Multiply out ( x)(x 4) ( x )(x 4) x 4x 6x 8 x x 8 You need to multiply each term in the first bracket by each term in the second. Use FOIL First, Outer, Inner, Last. You can see a step-by-step version of this example in the PowerPoint presentation Multiplying out brackets. You can also look at the Expanding brackets video, which shows multiplying out brackets, starting from an investigation that you might have seen a GCSE. It then looks at multiplying out first just one bracket, then two brackets. It then goes on to deal with some harder examples, which are beyond the requirements of this section but will be covered in the chapter on Polynomials, so are well worth looking at. The whole video lasts 40 minutes, so you may wish to fast-forward over parts of it if your time is limited. You can test yourself on multiplying out brackets using the Flash resource Expanding brackets. Factorising To factorise an expression, look for numbers and/or letters which are common factors of each term. We often talk about taking out a factor this MEI, 4/05/0 /8

3 MEI C Algebra Section Notes and Examples can cause confusion as it tends to make you think that subtraction is involved. In fact you are, of course, dividing each term by the common factor which you are taking out. Example 4 Factorise the following expressions. (i) 6a + b + c (ii) 6x²y 0xy² + xy (i) is a factor of each term. 6a + b + c = (a + 4b + c) (ii) xy is a factor of each term. 6x²y 0xy² + xy = xy(x 5y + ) You can check your answers by multiplying out the brackets. Multiplication Example 5 Simplify the expression xy yz 4x z. xy yz 4x z 4xx y yz z 4x y z You may be happy to do this in your head, without writing out the intermediate line of working. For practice in examples like this one, try the interactive questions Simplifying products. You may also like to try the Algebra Puzzle, either on your own or in a group, in which you need to be able to recognise equivalent algebraic expressions, involving simplifying, expanding brackets and multiplying. Algebraic fractions Algebraic fractions follow the same rules as numerical fractions. When adding or subtracting, you need to find the common denominator, which may be a number or an algebraic expression. MEI, 4/05/0 /8

4 MEI C Algebra Section Notes and Examples Example 6 Simplify x x 5x (i) 4 6 (ii) x x (i) The common denominator is, as, 4 and 6 are all factors of. x x 5x 8x x 0x 4 6 8x x 0x x (ii) The common denominator is x². x x x x x x x You are familiar with the idea of cancelling to simplify numerical fractions: for example, 9 can be simplified to 4 by dividing both the numerator and the denominator by. You can also cancel before carrying out a multiplication, to make the numbers simpler: e.g. 9. The same technique can be used in algebra. As with 4 factorising, remember that cancelling involves dividing, not subtracting. Example 7 Simplify 6xy x y (i) 0xy a a (ii) a a (i) It is very important to remember that you can only cancel if you can divide each term in both the numerator and denominator by the same expression. In this case, don t be tempted to divide by x²y although this is a factor of both x²y and 0x²y, it is not a factor of 6xy³. In a case like this, it may be best to factorise the top first, so that it is easier to see the factors. MEI, 4/05/0 4/8

5 MEI C Algebra Section Notes and Examples 6xy x y xy( y x) 0x y 0x y y x 5x (ii) Again, factorise where possible first. a a a ( a ) a a a a 6a a xy is a common factor of both top and bottom (a + ) is a common factor of both top and Notice that you cannot cancel a here, as it is not a factor of a +. You may also find the Mathcentre video Simplifying algebraic fractions useful. Linear equations A linear equation involves only terms in x (or whatever variable is being used) and numbers. So it has no terms involving x², x³ etc. Equations like these are called linear because the graph of an expression involving only terms in x and numbers (e.g. y = x + ) is always a straight line. Solving a linear equation may involve simple algebraic techniques such as gathering like terms and multiplying out brackets. Example 7 shows a variety of techniques that you might need to use. Example 8 Solve these equations. (i) 5x x 8 (ii) ( y) 4 ( y ) (iii) a a (i) 5x x 8 5x x 8 5xx0 5xx0 x 0 x 5 Add to each side Subtract x from each side Divide each side by MEI, 4/05/0 5/8

6 MEI C Algebra Section Notes and Examples (ii) ( y) 4 ( y ) 6y 4 y 6 6y 0 y 6y y 8y y 8 (iii) a a a (a ) a 6a 9 a 6a0 4a 0 a.5 Multiply out the brackets Add to each side Add y to each side Divide each side by 8 Multiply both sides by Multiply out the brackets Add to each side Subtract 6a from each side Divide both sides by -4 You can look at a demonstration of solving simple linear equations using the Flash resource Solving linear equations. You can also look at the Linear equations video which demonstrates the solution of a wide range of linear equations. In Example 9, the problem is given in words and you need to express this algebraically before solving the equation. Example 9 Sarah has a choice of two tariffs for text messages on her mobile phone. Tariff A: 0p for the first 5 messages each day, p for all others Tariff B: 4p per message How many messages would Sarah need to send each day for the two tariffs to cost the same? (She always sends at least 5!) Let the number of messages Sarah sends per day be n. Under Tariff A, she has to pay 0p for each of 5 messages and p for each of n - 5 messages. Cost = 50 ( n 5) Under Tariff B, she has to pay 4p for each of n messages. Cost = 4n For the cost to be the same MEI, 4/05/0 6/8

7 MEI C Algebra Section Notes and Examples 50 ( n 5) 4n 50 n0 4n 40 n4n 40 n 0 n She needs to send 0 messages per day for the two tariffs to cost the same. For practice in examples like this one, try the interactive questions Forming and solving linear equations. Changing the subject of a formula Changing the subject of a formula is similar to solving an equation, but you are working with letters rather than numbers. Example 0 The volume V of a cylinder with radius r and height h is given by V Make r the subject of this formula. r h. V r h V r h V r h V r h Divide both sides by h Square root both sides Finish by writing the equation with r on the left side. In the next example, two different methods are shown. The answers look a bit different but they are equivalent. Make sure that you can see how you could rewrite one solution to give the other. Example The surface area A of a cylinder with radius r and height h is given by A r( h r). Make h the subject of this formula. MEI, 4/05/0 7/8

8 MEI C Algebra Section Notes and Examples () A π r( h r) A πrh r A πr πrh A πr h πr A πr h πr () A π r( h r) A hr πr A r h πr A h r πr Multiply out the brackets Subtract r² from each side Divide each side by r Rewrite with h on the left side. Divide each side by r Subtract r from each side Rewrite with h on the left side In the next example, the new subject appears more than once. You need to collect the terms involving the new subject together and then factorise to isolate the new subject. Example Make x the subject of the formula cx a a( b x). cx a a( b x) cx a ab ax cx ax a ab cx ax ab a ( c a) x a( b ) ab ( ) x c a Multiply out the brackets Subtract ax from each side to collect the terms in x together Add a to each side Factorise Divide both sides by c - a The Rearranging formulae video shows a number of examples of rearranging formulae. MEI, 4/05/0 8/8

9 MEI Core Basic Algebra Section : Quadratics Notes and Examples These notes contain subsections on Factorising quadratic expressions Graphs of quadratic functions Solving quadratic equations by factorisation Solving quadratic equations using the formula Problem solving In this section of work you will be studying quadratic functions and their graphs. You probably already know something about this topic, but you will now be taking it a little further. Factorising quadratic expressions Again, this should be revision of GCSE work. It is essential that you are confident in factorisation. Example Factorise the expressions (i) x² + 4x + (ii) x² 4x (iii) x² 7x + 6 (i) x² + 4x + = (x..)(x...) x² + 4x + = (x + )(x + ) Start with an x in each bracket You need two numbers whose sum is 4 and whose product is. These are + and +. (ii) x² 4x = (x..)(x..) x² 4x = (x 6)(x + ) Start with an x in each bracket You need two numbers whose sum is 4 and whose product is. These are 6 and +. MEI, 6/0/09 /8

10 MEI C Algebra Section Notes and Examples (iii) x² 7x + 6 = (x..)(x..) x² 7x + 6 = (x )(x ) In this case you need to start with x in one bracket and x in the other. It is not so straightforward to find the two numbers in this case, because of the x in one bracket. The two numbers must have a product of +6, and as the coefficient of x is negative, they must both be negative. Try the different possibilities ( and 6, or and, in either order), until you find the correct one. You can see step-by-step examples of factorising quadratics in this PowerPoint presentation. You can also look at the Factorising quadratics video. For practice in examples like the ones above, try the interactive questions Factorising quadratics. You can also test yourself using the Flash resource Factorising quadratics. Sometimes algebraic expressions which look quite complicated can be simplified by factorising. Example x Simplify x x. x ( x )( x ) x x ( x )( x ) x x It can be tempting to try to cancel x² from the top and the bottom. Don t! You can only cancel something which is a factor of the top and the bottom. You can now cancel the factor (x ) from the top and the bottom. Graphs of quadratic functions Factorising a quadratic expression gives you information about the graph of a quadratic function. Do not think of this work as just algebraic manipulation, think about it also in terms of the graph of the function. Linking algebra and graphs is a very important mathematical skill; the good news is that being able to consider problems both algebraically and graphically usually makes them easier! A graphic calculator or computer package will be very useful. MEI, 6/0/09 /8

11 MEI C Algebra Section Notes and Examples You may already be familiar with the graph of the simplest quadratic function, y = x². y = x² The curve given by graphs of quadratics is called a parabola. Notice that all quadratic graphs have reflection symmetry. The mirror line is always a vertical line through the turning point or vertex of the curve (shown in yellow on these graphs). All other quadratic graphs have basically the same shape, but they may be stretched, squashed, shifted or inverted. y = x² x y = x x² Factorise the equations of these graphs. What is the relationship between the factorised form and the graph? Can you explain this? Notice that the graphs of functions with a negative x² term are inverted (upside down). You will look at the graphs of quadratic expressions in more detail in Polynomials Section. Solving quadratic equations by factorisation Solving quadratic equations is important not just from the algebraic point of view, but because it gives you information about the graph of a quadratic function. The solutions of the equation ax² + bx + c = 0 tells you where the graph of the function y = ax² + bx + c crosses the x-axis, since these are the points where y = 0. MEI, 6/0/09 /8

12 MEI C Algebra Section Notes and Examples Some quadratic equations can be solved by factorising. Example Solve these quadratic equations by factorising. (a) x² + x 8 = 0 (b) x² + x + = 0 (a) x² + x 8 = 0 (x + 4)(x ) = 0 x + 4 = 0 or x = 0 x = 4 or For this expression to be zero, one or other of the factors must be zero. (b) x² + x + = 0 (x + )(x + 4) = 0 x + = 0 or x + 4 = 0 x = or 4 You can see further examples using the Flash resources Quadratic equations (in which the coefficient of x² is always ) and Quadratic equations (in which the coefficient of x² is greater than ). For practice in examples like the ones above, try the interactive questions Solving quadratics by factorisation. Solving quadratic equations using the formula It is possible to solve quadratic equations using the method of completing the square (see pages 9 0 of the textbook); however the quadratic formula is just a generalisation of this method. Pages 0 of the textbook show how the quadratic formula is derived from the technique of completing the square. To help you understand this technique, you can look at the Flash resource Quadratic equations 4, and the Completing the square video. You may also find the Geogebra resource Completing the square useful this uses area to show what is happening when you complete the square. However, you should understand that completing the square is not normally used to solve a quadratic equation, since the quadratic formula is quicker and easier to use. Completing the square is covered in greater detail in Polynomials Section, where it is used to learn more about the graphs of quadratic functions. The quadratic formula for the solutions of the equation ax² bx c 0 is b b² 4ac x. a MEI, 6/0/09 4/8

13 MEI C Algebra Section Notes and Examples The expression b² 4ac is called the discriminant. This is very important as it tells you something about the nature of the solutions. In each case the solution(s) correspond to the points where the graph meets the x-axis. If the discriminant is positive, then there are two real solutions. (If the discriminant is a positive square number, then the two real solutions are rational and it is possible to solve the equation by factorisation; otherwise the solutions are irrational and you must use the quadratic formula.) y = x² + x Discriminant = 6 Two rational solutions y = x² + x Discriminant = Two real, irrational solutions If the discriminant is zero, then the quadratic is a perfect square and there is one real solution, which can be found by factorisation. y = x² + x + Discriminant = 0 One real solution If the discriminant is negative, then there are no real solutions. y = x² + x + Discriminant = -4 No real solutions As the graph does not meet the x-axis, there cannot be any real solutions. MEI, 6/0/09 5/8

14 MEI C Algebra Section Notes and Examples When you need to solve a quadratic equation, it is useful to quickly work out the discriminant before you start, so that you know whether there are real solutions, and whether the equation can be solved by factorisation. Note that when you use the quadratic formula in C, you will need to give answers in an exact form, involving a square root, since C is a noncalculator paper. Sometimes you may need to simplify square roots, e.g. 8 4, so that you can cancel down fractions. You will learn more about working with square roots in chapter 5. You can look at the relationship between the discriminant of a quadratic equation and the number of roots using the Flash resource The discriminant. You can also see examples of solving quadratic equations using the formula in the Flash resource Quadratic equations. The Solving quadratics video looks at solving quadratics by all the methods covered. Try the Quadratics puzzle, either on your own or with one or two others. Cut out all the pieces and match up each equation with its solution. The pieces will form a large hexagon. Example 4 For each of the following quadratic equations, find the discriminant and solve the equation, where possible, by a suitable method (i) x ² 5x 0 (ii) 6x ² x 0 0 (iii) x ² x 4 0 (iv) 4x ² x 9 0 (i) a =, b = -5, c = Discriminant = ( 5)² Since the discriminant is positive, there are two real solutions. As it is not a square number, the equation must be solved using the quadratic formula. b b² 4ac x a (ii) a = 6, b =, c = -0 Discriminant = ² Since the discriminant is positive, there are two real solutions. As it is a square MEI, 6/0/09 6/8

15 MEI C Algebra Section Notes and Examples number (9²), the equation can be solved by factorisation. 6x² x 0 0 (x )(x 5) 0 x or x 5 (iii) a =, b = -, c = 4 Discriminant = ( )² Since the discriminant is negative, there are no real solutions. (iv) a = 4, b =, c = 9 Discriminant = ² Since the discriminant is zero, there is one solution and the equation can be solved by factorisation into a perfect square. 4x² x 9 0 (x )² 0 x Problem solving Some problems, when translated into algebra, involve quadratic equations. Example 5 A rectangular box has width cm greater than its length, and height cm less than its length. The total surface area of the box is 548 cm². What are the dimensions of the box? Let the length of the box be x cm. The width of the box is x + cm, and the height is x cm. The surface are of the box is given by x(x + ) + x(x ) + (x + )(x ) x(x + ) + x(x ) + (x + )(x ) = 548 x(x + ) + x(x ) + (x + )(x ) = 74 x² + x + x² x + x² x 6 = 74 x² x 80 = 0 (x + 8)(x 0) = 0 x = 0 Divide through by The discriminant is 64, which is 58², so this must factorise x + 8 = 0 gives a negative value of x, which does not make sense in this context. So the solution must be x 0 = 0. The length of the box is 0 cm, the width is cm and the height is 7 cm. MEI, 6/0/09 7/8

16 MEI C Algebra Section Notes and Examples Notice that in Example 5, you could discard one of the possible solutions as a negative solution did not make sense in the context. This is not always the case. In some situations, a negative solution can have a practical meaning. For example if the height of a stone thrown from the edge of a cliff is negative, this simply means that the stone is below the level of the cliff at that point. However, if the stone was thrown from level ground, then a negative height does not make sense. Some problems leading to quadratic equations do have two possible solutions. Always consider whether your solution(s) make sense in the context. For practice in examples like the ones above, try the interactive questions Forming and solving quadratics. MEI, 6/0/09 8/8

17 MEI Core Basic Algebra Section : Simultaneous equations Notes and Examples These notes contain subsections on Linear simultaneous equations One linear and one quadratic equation Linear simultaneous equations Simultaneous equations involve more than one equation and more than one unknown. To solve them you need the same number of equations as there are unknowns. One method of solving simultaneous equations involves adding or subtracting multiples of the two equations so that one unknown disappears. This method is called elimination, and is shown in the next example. Example Solve the simultaneous equations pq5 pq4 Adding or subtracting these equations will not eliminate either p or q. However, you can multiply the first equation by, and then add. This will eliminate p. pq5 pq4 6 pq 0 pq 4 Adding: 7 p 4 p q 5 6q 5 q The solution is p =, q = -. Now substitute this value for p into one of the original equations you can use either, but in this example, is used. MEI, 6/0/09 /4

18 MEI C Algebra Section Notes and Examples Notice that, in Example, you could have multiplied equation by and then subtracted. This would give the same answer. Sometimes you need to multiply each equation by a different number before you can add or subtract. This is the case in the next example. Example 5 pencils and rubbers cost.50 8 pencils and rubbers cost.5 Find the cost of a pencil and the cost of a rubber. 5pr 50 8pr 5 Let p represent the cost of a pencil and r represent the cost of a rubber. It is easier to work in pence. The easiest method is to multiply equation by and equation by. (You could of course multiply by 8 and by 5). 5 p6r p6r 470 Subtracting: p 0 p r r 50 r 50 r 5 A pencil costs 0p and a rubber costs 5p. Substitute this value of p into equation An alternative method of solving simultaneous equations is called substitution. This can be the easier method to use in cases where one equation gives one of the variables in terms of the other. This is shown in the next example. Example Solve the simultaneous equations xy y 5x MEI, 6/0/09 /4

19 MEI C Algebra Section Notes and Examples x (5 x) x0 4x 7x x y 5 56 Substitute the expression for y given in the second equation, into the first equation. Multiply out the brackets Substitute the value for x into the original second equation The solution is x =, y = - For some practice in examples like the ones above, try the interactive questions Solving linear simultaneous equations and Forming and solving linear simultaneous equations. You can also look at the Simultaneous equations video. One linear and one quadratic equation When you need to solve a pair of simultaneous equations, one of which is linear and one of which is quadratic, you need to substitute the linear equation into the quadratic equation. Example 4 Solve the simultaneous equations x y 6 x y x y ( y) y 6 y y y 6 y y 5 0 (y 5)( y) 0 y 5 or y Start by using the linear equation to write one variable in terms of the other. Now substitute this expression for y into the first equation Multiply out, simplify and factorise Sometimes you will need to use the quadratic formula to solve the resulting quadratic equation. 5 5 y x y y x y 5 The solutions are x, y and x, y Now substitute each value for y into the linear equation to find the corresponding values of x MEI, 6/0/09 /4

20 MEI C Algebra Section Notes and Examples When you try to solve a quadratic equation and a linear equation simultaneously, you usually get solutions. However, 0 or solutions are also possible. Can you explain why? [Hint: think about the graphs]. You can look at some more examples like these using the Flash resource Simultaneous equations. For some practice in examples like the ones above, involving a quadratic equation, try the interactive resource Simultaneous equations involving quadratics. MEI, 6/0/09 4/4

21 MEI Core Coordinate Geometry Section : Points and straight lines Notes and Examples These notes contain sub-sections on: Gradients, distances and mid-points The equation of a straight line The intersection of two lines Gradients, distances and mid-points You will have met gradients before at GCSE. To revise finding the gradient of a line from a diagram, use the interactive questions The gradient of a line. Remember that lines which go downhill have negative gradients. To find the gradient of a straight line between two points use the formula y y gradient. x x x y and, If two lines are parallel, they have the same gradient. If two lines with gradients m and m are perpendicular, then mm, x y, Example P is the point (-, 7). Q is the point (5, ). Calculate (i) the gradient of PQ (ii) the gradient of a line parallel to PQ (iii) the gradient of a line perpendicular to PQ. (i) Choose P as ( x, y ) and Q as ( x, y ). or vice versa: it will still give the same answer (WHY?) Gradient of PQ = y x y x = 7 5 ( ) = 6 8 = 4 Notes: () Draw a sketch and check that your answer is sensible (e.g. has negative gradient). () Check that you get the same result when you choose Q as ( x, y ) and P as ( x, y ). MEI, 4/05/0 /7

22 MEI C Coordinate geom. Section Notes & Examples (ii) When two lines are parallel their gradients are equal. ( m = m ) So the gradient of the line parallel to PQ is also. 4 (iii) When two lines are perpendicular mm. So m 4 4 m The gradient of a line perpendicular to PQ is 4. For more examples on gradient, look at the Flash resource Gradient of a line. You may also find the Mathcentre video The gradient of a straight line segment useful. For further practice in examples like the one above, try the interactive questions The gradient of a line between two points, Collinear points and The gradient of a perpendicular. The midpoint of a line joining two points Midpoint x x, y y x y and,, x y is given by x y, The x-coordinate of M is halfway between x and x. The y-coordinate of M is halfway between y and y. M x y, The length of a line joining two points Pythagoras Theorem. x y and,, x y can be found using Length x x y y x y, y y x y, x x MEI, 4/05/0 /7

23 MEI C Coordinate geom. Section Notes & Examples Example A is the point (, -6). B is the point (-, 4). Calculate (i) the midpoint of AB (ii) the length of AB. or vice versa, it will still give the same answer (WHY?) Choose A as ( x, y ) and B as ( x, y ). (i) Midpoint is x x y y, i.e. ( ) 6 4, =, (ii) The distance AB is given by d ( x x ) ( y y ) ( ( )) (( 6) 4) (5) ( 0) Note: The answer is often left like this if the square root is not exact. However since 5 = 5 5 then is perhaps a simpler form. To see more examples like these, try the Flash resources Distance between two points and Midpoint of two points. You may also find the Mathcentre video Properties of straight line segments useful. For further practice in examples like the one above, try the interactive questions The distance between two points and The midpoint between two points. The equation of a straight line The equation of a straight line is often written in the form y = mx + c, where m is the gradient and c is the intercept with the y-axis. Example Find (i) the gradient and (ii) the y-intercept of the following straight-line equations. (a) 5y7x (b) x8y7 0 MEI, 4/05/0 /7

24 MEI C Coordinate geom. Section Notes & Examples (a) Rearrange the equation into the form y = mx + c. 7 5y7x becomes y x so m = 7 and 5 c 5 (i) The gradient is 7 5 (ii) The y-intercept is Note the minus sign (b) Rearrange the equation into the form y = mx + c. x8y7 0 becomes 8y x 7 giving y = 8 x+ 7 8 so m = 8 and c = 7 8 (i) The gradient is 8 (ii) The y-intercept is 7. 8 Note the minus sign Sometimes you may need to sketch the graph of a line. A sketch is a simple diagram showing the line in relation to the origin. It should also show the coordinates of the points where it cuts one or both axes. You can explore straight line graphs using the Flash resources Equation of a line y = mx + c and Equation of a line ax + by + c = 0. You may also find the Mathcentre video Equations of a straight line and Linear functions and graphs useful. Example 4 Sketch the lines (a) 5y7x (b) x8y7 0 (a) From Example you know that 5y7x has gradient 7 and y-intercept 5. 5 Sketch of 5y7x The gradient is positive, so the line slopes upwards from left to right. It s also greater than and so steeper than 45 degrees. y This means the line goes through (0, -0.6) which is below the origin. 5 x MEI, 4/05/0 4/7

25 MEI C Coordinate geom. Section Notes & Examples (b) From Example you know that x8y7 0 has gradient and y-intercept The gradient is negative, so the line slopes downwards from left to right. It is also and less than and so less steep than 45 degrees. This means the line goes through (0, 0.875) which is above the origin. Sketch of x8y7 0 y 7 8 x Sometimes you may need to find the equation of a line given certain information about it. If you are given the gradient and intercept, this is easy: you can simply use the form y = mx + c. However, more often you will be given the information in a different form, such as the gradient of the line and the coordinates of one point on the line (as in Example 5) or just the coordinates of two points on the line (as in Example 6). In such cases you can use the alternative form of the equation of a straight line. For a line with gradient m passing through the point x, y, the equation of the line is given by y y m x x. Example 5 (i) Find the equation of the line with gradient and passing through (, -). (ii) Find the equation of the line perpendicular to the line in (i) and passing through (, -). (i) The equation of the line is y y mx x y x y x 6 y x 7 m = and ( x, y ) is (, -) You should check that the point (, -) satisfies your line. If it doesn t, you must have made a mistake! MEI, 4/05/0 5/7

26 MEI C Coordinate geom. Section Notes & Examples (ii) For two perpendicular lines mm, so the gradient of the new line is. The equation of the line is y y mx x y x y x y x y x m and x y is (, -), The final form of the equation can be written in various different ways: e.g. y = -x + (This form has no fractions.) e.g. y + x = (This has no fractions and avoids having a negative sign at the start of the right hand side.) You can see more examples like this using the Flash resource Equation of a line y y =m(x x). Also look at Parallel lines and Perpendicular lines. In the next example, you are given the coordinates of two points on the line. Example 6 P is the point (, 8). Q is the point (-, 5). Find the equation of PQ. Choose P as ( x, y ) and Q as ( x, y ). One method is to find the gradient and then use this value and one of the points in y y mx x Gradient of PQ = y x y x 5 8 = = 4 = 4 Now use y y mx x y8 4 x 4 y8 x 4y x 9 4y x You should check that P and Q satisfy your line. MEI, 4/05/0 6/7

27 MEI C Coordinate geom. Section Notes & Examples An alternative approach to the above examples is to put the formula for m into the straight line equation to obtain y y y y x x x x and then make the substitutions. This is equivalent to the first method, but does not involve calculating m separately first. For further practice in examples like the one above, try the interactive questions The equation of a line between two points. The Lines Activities include a variety of activities which involve the equations of lines and their properties. The intersection of two lines The point of intersection of two lines is found by solving the equations of the lines simultaneously. This can be done in a variety of ways. When both equations are given in the form y = then equating the right hand sides is a good approach (see below). If both equations are not in this form, you can rearrange them into this form first, then apply the same method. Alternatively, you can use the elimination method (see Basic Algebra section ) if the equations are in an appropriate form. Example 7 Find the point of intersection of the lines yxand y5x 8. x 5x8 x 8 6x x Substitute into one of the equations to find y Substituting x = into yx gives y 7 The point of intersection is (, 7) Check that (, 7) satisfies the second equation. You can see more examples like this using the Flash resource Intersection of two lines. MEI, 4/05/0 7/7

28 MEI Core Coordinate Geometry Section : Curves and circles Notes and Examples These notes and examples contain subsections on The equation of a circle Finding the equation of a circle Circle geometry The intersection of a line and a curve The intersection of two curves The equation of a circle Start this section by looking at either the Geogebra resource Circles or the Circles dynamic spreadsheet (selecting the Circle Equations sheet). First, set the centre of the circle to be the origin and vary the radius. Look at how the equation of the circle changes. Now vary the coordinates of the centre of the circle, and look at how the equation of the circle changes. You can also explore equations of circles using the Flash resources Equation of a circle centre O and Equation of a circle centre (a, b). You should find out the following results, which you need to learn: The general equation of a circle, centre the origin and radius r is x y r The general equation of a circle, centre (a, b) and radius r is ( x a) ( y b) r Make sure you understand why these equations describe circles. See page 6 in the textbook for help. Example For each of the following circles find (i) the coordinates of the centre and (ii) the radius. (a) x² + y² = 49 (b) (x + )² + (y 6)² = 9 MEI, 0/0/ /9

29 MEI C Coordinate geom. Section Notes & Examples (a) x² + y² = 49 can be written as x² + y² = 7². (i) The coordinates of the centre are (0, 0) (ii) The radius is 7. This is a particular case of the general form x² + y² = r² which has centre (0, 0) and radius r. (b) (x + )² + (y 6)² = 9 can be written as (x ())² + (y 6)² = ². (i) The coordinates of the centre are (, 6) (ii) The radius is. This is a particular case of the general form (x a)² + (y b)² = r² which has centre (a, b) and radius r. For practice in examples like the one above, try the interactive questions Finding the radius and centre of a circle (circle equation in its simplest form). The first activity in Circle Activities gives you practice in matching equations of circles with diagrams. Sometimes the circle equation needs to be rearranged into its standard form before you can find the centre and radius. Example Show that the equation and radius. x y x y represents a circle, and find its centre The general equation of a circle is Multiplying out: ( x a) ( y b) r x ax a y by b r x y ax by a b r 0 Comparing with the original equation: a 4 a The equation can be written as ( x ) ( y) 4 This is the equation of a circle, centre (-, ), radius 4. b 6 b a b r 4 9 r r 6 MEI, 0/0/ /9

30 MEI C Coordinate geom. Section Notes & Examples For practice in examples like the one above, try the interactive questions Finding the radius and centre of a circle (circle equation in its expanded form). In the example above, you are using the technique of completing the square, which is covered briefly in Chapter (pages 9 0), and in more depth in Chapter (pages 98 99). You may find the Mathcentre video Coordinate geometry of a circle useful. Finding the equation of a circle In Section you looked at different ways of finding the equation of a line. You can find the equation of a line from the gradient and the intercept, or from the gradient and one point on the line, or from two points on the line. In the same way, there are several ways of finding the equation of a circle, depending on the information available. Finding the equation of a circle from the radius and centre Example Find the equation of each of the following. (a) a circle, centre (0, 0) and radius 4. (b) a circle, centre (, 4) and radius 6. (a) The equation of a circle centre the origin is x² + y² = r² r = 4 so the equation is x² + y² = 4² i.e. x² + y² = 6 (b) The equation of a circle centre (a, b) and radius r is (x a)² + (y b)² = r² a =, b = 4 and r = 6 so the equation is (x )² + (y (4))² = 6² i.e. ( x)² ( y 4)² 6 Finding the equation of a circle from its centre and one point on its circumference If you know the centre of the circle and one point on its circumference, you can find the radius by calculating the distance between these two points. You can then find the equation of the circle. Example 4 Find the equation of the circle, centre (, -), which passes through the point (-, -). MEI, 0/0/ /9

31 MEI C Coordinate geom. Section Notes & Examples The distance r between (, -) and (-, -) is given by: ( ) ( ) 0 r The radius of the circle is therefore 0. The equation of the circle is ( x) ( y ) 0 For practice in examples like the one above, try the interactive questions Find the equation of a circle. Finding the equation of a circle from three points on its circumference To find the equation of a line, you need the coordinates of two points on the line. To find the equation of a circle, you need the coordinates of three points on the circumference of the circle. One method is illustrated by the Circles dynamic spreadsheet. Select the sheet Circumcentre and follow the instructions on the sheet. This demonstration shows that the centre of the circle is the intersection of the perpendicular bisector of each pair of points. To find the centre of a circle through three points A, B and C, it is sufficient to find two of the perpendicular bisectors. For example, you can find the equations of the perpendicular bisectors of AB and BC, and then solve these equations simultaneously to find the point of intersection, i.e. the centre of the circle. You can then use the coordinates of the centre and one of the three points A, B and C to find the radius of the circle (as in Example 4), and hence find the equation of the circle. Example 5 Find the equation of the circle passing through A (, ), B (9, ) and C (8, 4). C (8, 4) B (9, ) A sketch is often helpful. The sketch does not need to be accurate. It gives some idea of roughly where the centre is, so you can check your answer is reasonable. A (, -) MEI, 0/0/ 4/9

32 MEI C Coordinate geom. Section Notes & Examples You want to find the equation of the perpendicular bisector of AB. This is perpendicular to AB and passes through the midpoint M of AB. The gradient of AB is found by using m y y x x m ( ) 4 = 9 8 = Note: Looking at the sketch we expect the gradient of AB to be positive. Using mm, the gradient of the perpendicular bisector is. The midpoint M of AB is found by using M = x x, y y. 9 You are given A(, -) and B(9, ) so M is, = (5, ) The perpendicular bisector is found using y y m( x x ) with ( x, y ) = (5, ) and m =. so y ( ) ( x 5 ) y + = x + 0 y = x + 9 (equation I) Next, use the same method to find the perpendicular bisector of BC. The gradient of BC is Note: Looking at the sketch, we expect the gradient of BC to be negative. Therefore the gradient of the perpendicular bisector of BC is The midpoint N of BC is, so N is (8.5,.5). The equation of the perpendicular bisector is y.5 = (x 8.5) (y.5) = x 8.5 y 7.5 = x 8.5 y = x (equation II) y = x + 9 (equation I) y = x (equation II) Next, find the coordinates of the centre of the circle by solving equations (I) and (II) simultaneously. MEI, 0/0/ 5/9

33 MEI C Coordinate geom. Section Notes & Examples Substituting (I) into (II) (x + 9) = x 6x + 7 = x 8 = 7x x = 4 Substituting x = 4 into equation (I) gives y = (4) + 9 = So the coordinates of the centre are (4, ). Note: Looking at the sketch this appears to be a plausible result. The radius is the distance between the centre (4, ) and a point on the circumference such as (9, ). This can be found by using d ( x x ) ( y y ) radius = ( 9 4) ( ) = 5 = 5 Finally, using the general form (x a)² + (y b)² = r² with a = 4, b = and r = 5 the equation of the circle is (x 4)² + (y )² = 5. Note: You should check that each of the points A, B and C satisfy this equation. Circle geometry The three facts about circles given on pages 6 and 64 are important. They often help to solve problems involving circles.. The angle in a semicircle is a right angle. See the Flash resource The angle in a semicircle for a demonstration.. The perpendicular from the centre of a circle to a chord bisects the chord. See the Flash resource Perpendicular to a chord for a demonstration.. The tangent to a circle is perpendicular to the radius at that point See the Flash resource Tangent and radius for a demonstration. Keep these properties in mind when dealing with problems involving circles. For some practice in using the third property, try the interactive questions Tangents and normals to circles.. The second activity in Circles activities looks at using the third property to find the equation of the tangent to a circle. MEI, 0/0/ 6/9

34 MEI C Coordinate geom. Section Notes & Examples The intersection of a line and a curve Just as the point of intersection of two straight lines can be found by solving the equations of the two lines simultaneously, the point(s) of intersection of a line and a curve can be found by solving their equations simultaneously. In many cases, the equations of both the line and the curve are given as an expression for y in terms of x. When this is the case, a sensible first step is to equate the expressions for y, as this leads to an equation in x only. Example 6 Find the coordinates of the points where the line y = x + meets the curve y = x² x + 5. x² x + 5 = x + x² 4x + = 0 (x )(x ) = 0 x = or x = When x = then y = + = 5 When x = then y = + =. Equate the expressions for y to give an equation in x only Substitute the x values into the equation of the line The points where the line meets the curve are (, 5) and (, ). You should check that each of these points satisfies the equation of the curve. (You have already used the equation of the line to find the y-values). Notice that this problem involved solving a quadratic equation, which in this case had two solutions, showing that the line crossed the curve twice. However, the quadratic equation could have had no solutions, which would indicate that the line did not meet the curve at all, or one repeated solution, which would indicate that the line touches the curve. You can look at some examples with the Flash resource Intersection of a curve and a line. For practice in examples like the one above, try the interactive questions Quadratic and line intersection. The next example looks at the intersection of a line and a circle. Before reading this example, look at the Circles dynamic spreadsheet. Select the sheet Circle and a line. Try varying the equation of the line and/or the circle, MEI, 0/0/ 7/9

35 MEI C Coordinate geom. Section Notes & Examples and make sure that you can see that there may be two intersections, no intersections or one intersection (in which case the line touches the circle). You can also look at the Flash resource Intersection of a circle and a line. Example 7 Find the coordinates of the point(s) where the circle (i) the line y = 5 (ii) the line x = (iii) the line y = x ( x ) ( y ) 9 meets (i) Substituting y = 5 into the equation of the circle: ( x ) (5 ) 9 ( x ) 6 9 ( x ) 7 There are no solutions. The line does not meet the circle. The expression (x + )² cannot be negative (ii) (iii) Substituting x = into the equation of the circle: ( ) ( y ) 9 9 ( y ) 9 ( y ) 0 y The line touches the circle at (, ). Substituting y = x into the equation of the circle: ( x ) ( x) 9 ( x ) ( x) 9 x 4x 4 x x 9 x x 4 0 x x 0 ( x)( x ) 0 x or x The point is on the line x =, so its x-coordinate must be. Substitute the x values into the equation of the line to find the y-coordinates. When x =, y = = When x =, y = ( ) = 4 The line crosses the circle at (, ) and (, 4). For practice in examples like the one above, try the interactive questions Circle and line intersection. MEI, 0/0/ 8/9

36 MEI C Coordinate geom. Section Notes & Examples The intersection of two curves As before, the intersections of two curves can be found by solving the equations of the curves simultaneously. As in Example 4, in many cases a sensible first step is to equate the expressions for y. Example 8 Find the coordinates of the points where the curve y = x² 6x + 5 intersects the curve y = x² + x 9. x² 6x + 5 = x² + x 9 x² 8x + 4 = 0 x² 6x + 8 = 0 (x )(x 4) = 0 x = or x = 4 Equate the expressions for y to give an equation in x only There is a common factor of, so divide by before factorising. When x =, y = ()² - 6() + 5 = When x = 4, y = (4)² 6(4) + 5 = The points of intersection are (, ) and (4, ). Substitute the x values into the equation of one of the curves (in this case the first one). You should check that each of these points satisfy the equation of the second curve. (You have already used the equation of the first curve to find the y-values). MEI, 0/0/ 9/9

37 Notes and Examples MEI Core Polynomials Section : Introducing Polynomials These notes contain subsections on Adding and subtracting polynomials Multiplying polynomials Dividing polynomials Graphs of polynomial functions Sketching graphs of polynomials in factorised form Finding the equation of a curve This section involves quite a lot of algebraic manipulation. It may look a bit daunting, but you will probably find that it is easier than it looks once you have got the hang of it! There are a lot of possible different approaches to the examples in this section. You may also see different variations in your textbook or in class, or you may find your own method. In these notes, the examples are done using fairly traditional methods, as these are easy to show on a printed page. Other approaches are difficult to show on a printed page because of the thought processes involved, so you will also find links to PowerPoint presentations that demonstrate these methods step-by-step. You will find several in your head methods in the PowerPoint presentations. You may find these approaches difficult at first. If you do, then leave them and choose a different method, but it is well worth returning to them and trying again later in your course as you become more confident. Adding and subtracting polynomials Example For the polynomials f(x) = x³ x² + g(x) = x³ + x² x 4 find (i) f(x) + g(x) (ii) f(x) g(x) (i) (x³ x² + ) + (x³ + x² x 4) = (x³ + x³) + (-x² + x²) + (-x) + ( 4) = x³ x² x Add the terms in x³, the terms in x², the terms in x and the constant terms separately. Notice that f(x) does not have a term in x. It is not essential to put in the brackets as shown here, but it can be helpful. MEI, 4/05/0 /9

38 MEI C Polynomials Section Notes and Examples + x x x x x 4 x x x Alternatively you can write it out like an addition sum: (ii) (x³ x² + ) (x³ + x² x 4) = (x³ x³) + (-x² x²) + (x) + ( + 4) = x³ 4x² + x + 5 Do this in a similar way, but be careful about signs. x x x x x x x x Alternatively (again be careful about signs): You can see more examples like this using the Flash resources Polynomial addition and Polynomial subtraction. Multiplying polynomials You are already familiar with multiplying out two linear expressions to obtain a quadratic expression. You now need to be able to deal with more complicated multiplications. The principle is the same for a multiplication involving two brackets, each term in one bracket needs to be multiplied by each term in the other bracket. First of all, it is helpful to think about how many terms there will be in the result (before simplifying). Each of the terms in the first bracket must be multiplied by each of the terms in the second bracket, so there should be 9 terms altogether. There are a number of different ways of setting out this multiplication. One way is shown in Example below. You can also view the animated PowerPoint presentation Multiplying polynomials, which shows two other methods: () using a table, and () doing it in your head. Example Multiply x² + x by x² x + 4 Each term in the first bracket is multiplied by the whole of the second bracket (x² + x )(x² x + 4) = x²(x² x + 4) + x(x² x + 4) (x² x + 4) = (x 4 x³ + 4x²) + (6x³ x² + x) + (-4x² + x 8) 9 terms as expected MEI, 4/05/0 /9

39 MEI C Polynomials Section Notes and Examples = x 4 + ( x³ + 6x³) + (4x² x² 4x²) + (x + x) 8 = x 4 + 5x³ x² + 4x 8 Collect like terms together You can see more examples using the Flash resource Multiplying polynomials. You also need to be able to multiply out expressions involving more than one bracket. This is shown in the next example. Example Multiply out (x )(x + )(x ) It is often easiest to multiply out one pair of brackets, and then multiply the result by the third bracket. (x )(x + ) = x² + x 4x 6 = x² x 6 (x² x 6)(x ) = x²(x ) x(x ) 6(x ) = 6x³ x² x² + x 8x + 6 = 6x³ 5x² 7x + 6 Simplify here or the next step will be more complicated! You can of course use another approach, such as using a table, or doing it in your head, to multiply the quadratic by the third bracket. You can also do the whole thing in your head. The PowerPoint presentation Multiplying three brackets gives an animated example. Dividing polynomials When you divide one polynomial by another, it may divide exactly, or there may be a remainder, just as in arithmetic. For example: 6 6 = 4 remainder You could rewrite this statement as: 6 is called the dividend, 6 is called the divisor, 4 is the quotient and is the remainder. 6 = This rearrangement helps to give one method of dividing polynomials. When working with polynomials, remember the following points: If you are dividing by a linear expression, the quotient is of order one less than the dividend (e.g. for a quartic, the quotient is cubic) and the remainder, if any, is a constant term. MEI, 4/05/0 /9

40 MEI C Polynomials Section Notes and Examples If you are dividing by a quadratic term, the quotient is of order two less than the dividend (e.g. for a quartic, the quotient is quadratic) and the remainder, if any, could be linear or a constant term. This idea can be extended to a polynomial divisor of any order. Most, if not all, of the examples you meet will only involve dividing by a linear expression. Example 4 Divide x³ + x² - x + by x + When you divide a cubic expression by a linear expression, as in this example, the quotient is a quadratic expression and the remainder, if any, is a constant term. Let the quotient be ax² + bx + c and the remainder be d. x³ + x² x + = (x + )(ax² + bx + c) + d = x(ax² + bx + c) + (ax² + bx + ) + d = ax³ + bx² + cx + ax² + bx + c + d = ax³ + (b + a)x² + (c + b)x + c + d Equating coefficients of x³ a = Equating coefficients of x² b + a = b + 4 = b = - Equating coefficients of x c + b = - c = - c = Equating constant terms c + d = + d = d = - x³ + x² x + = (x + )(x² x + ) The quotient is x² x + and the remainder is -. Don t be put off by a negative remainder, as in this example this is quite acceptable in polynomial division! There are other approaches to polynomial division. The PowerPoint presentation Dividing polynomials shows () algebraic long division and () polynomial division by inspection (i.e. in your head). The Geogebra resource Polynomial division shows the box method. You can also look at more examples using the Flash resources Polynomial division (long division), Polynomial division by inspection, Polynomial division: the box method (no remainder) and Polynomial division: the box method (remainder). There is also a Polynomial division video this deals with the long division method only. For some extra practice in examples like the one above, try the interactive resource Dividing polynomials. MEI, 4/05/0 4/9

41 MEI C Polynomials Section Notes and Examples Graphs of polynomial functions There are two simple rules about the graphs of polynomial functions: a polynomial of order n meets the x-axis at most n times. a polynomial of order n has at most n turning points. For example, a quadratic, which has order, has turning point and meets the x-axis at most twice. This quadratic crosses the x-axis twice. This quadratic touches the x-axis once. This quadratic does not meet the x-axis at all. A cubic, which has order, has at most turning points and meets the x-axis at most three times and at least once. Here are some examples. This cubic graph has two turning points and crosses the x-axis three times. This cubic graph has two turning points, but only crosses the x-axis once. The simplest cubic graph of all, y = x³, has just one turning point, a point of inflection. You can think of this as two turning points in the same place. It crosses the x-axis just once. A quartic, which has order 4, has at most turning points and meets the x-axis at most four times. Here are some examples. MEI, 4/05/0 5/9

42 MEI C Polynomials Section Notes and Examples This quartic graph has three turning points and crosses the x-axis four times. This quartic graph has three turning points and crosses the x-axis twice. This quartic graph has just one turning point and crosses the x-axis twice. Notice that the turning point is much flatter than usual it is actually three turning points all in the same place. This makes the shape different from a quadratic graph. Sketching graphs of polynomials in factorised form You have already done some work on factorising quadratic expressions. As you know, a quadratic expression can sometimes be factorised into two linear factors. These factors can be used to tell you where the graph of the quadratic cuts the x-axis. For example, the quadratic expression y = x² + x can be written in factorised form as y = (x )(x + ) When the graph cuts the x-axis, the value of y = 0. (x )(x + ) = 0 x = or x = -. The graph therefore cuts the x-axis at (, 0) and (-, 0). You can also find out where the graph cuts the y-axis by substituting x = 0. In this case, when x = 0, y = -, so the graph cuts the y-axis at (0, -). This information allows you to sketch the graph. y = x² + x MEI, 4/05/0 6/9

43 MEI C Polynomials Section Notes and Examples It is also useful to think about the behaviour of the graph as x ( i.e. when x is very large and positive) and as x - ( i.e. when x is very large and negative). For the quadratic graph above, for values of x which are numerically large the term x² is the most significant term (the dominant term). So you just need to think about whether this term is positive or negative. In this case it is positive as x and as x -. So the graph disappears off the top of the page at both the left and right of the graph. The same ideas can be extended to any polynomial. You will learn how to factorise cubics and higher order polynomials in Section. For now, you will look at polynomials that are given in factorised form. Examination questions commonly ask you to sketch the graph of a function which you have already factorised. This should be an easy source of marks (provided you have managed the factorising!) but a lot of candidates throw marks away. Here are some tips: A sketch means a sketch! Do it in the answer booklet, not on graph paper (just because graph paper is provided does not mean that you are expected to use it!). You should certainly not be calculating and plotting points this is very time-consuming, unnecessary and may result in only part of the graph. Show the points at which the graph cuts BOTH axes. The points at which the graph cuts the x-axis can be found from the factorised equation. Find the point at which the graph cuts the y-axis by substituting x = 0, and mark on this point one mark is often given for this. Make sure that your graph is the right way up. If you have marked on the point where the graph cuts the y-axis then it should be clear how the graph goes. Also think about whether y is positive or negative when x is very large and positive, and when x is very large and negative. Do not stop the graph when you reach an axis! You may lose marks for this. The graph should go beyond each of the points marked. Example 5 Sketch the following graphs, showing the points where the graphs meet the coordinate axes. (i) y ( x )( x )(x ) (ii) y x( x ) ( x ) (i) ( )( )( ) y x x x When y = 0, x =, x = - or x =. MEI, 4/05/0 7/9

44 MEI C Polynomials Section Notes and Examples When x = 0, y = - = - The graph crosses the x-axis at (, 0), (-, 0) and (, 0). The graph crosses the y-axis at (0, -). y ( x )( x )(x ) As x, y is positive. - - As x -, y is negative. (ii) y x( x ) ( x ) When y = 0, x = 0, x = (repeated) or x = -. When a root is repeated, the graph touches the x-axis There is no need to look at where the graph crosses the y-axis, since you already know it passes through the origin. The graph crosses the x-axis at (0, 0) and (-, 0), and touches the x-axis at (, 0). - 0 As x, y is positive. As x -, y is positive. For some examples, look at the Flash resource Sketching factorised cubics. For each example, look at the equation in its factorised form and work out the points at which the graph will cross both axes, and decide which way up it will be. Then press Show curve to check. You may also find the Mathcentre video Polynomial functions useful. For some extra practice in recognising the shapes of quadratic and curves, try the interactive questions Sketching polynomial curves. (If you look at the worked solutions, don t worry about the mention of differentiation which is MEI, 4/05/0 8/9

45 MEI C Polynomials Section Notes and Examples covered in Core all you need to know at this stage is that quadratics have one turning point and that cubics can have up to two). Finding the equation of a curve If you know where a polynomial curve cuts the axes (or alternatively, if you are given the roots of a polynomial equation), you can deduce the equation of the curve by writing it in factorised form and then multiplying out the brackets. Example 6 Find a polynomial equation which has roots x =, x = - and x = 0.5. A polynomial equation with these roots is ( x )( x )(x ) 0. Multiplying out: An equation with these roots is ( x )( x )(x ) ( x x )(x ) x x 8x 0 x 4x 6x x x x x 8x Note that any multiple of this polynomial (e.g. 4x 6x 8x 6 0) would have the same roots. This is the simplest possible equation. For some extra practice in examples like the one above, try the interactive resource Finding a polynomial from its roots. MEI, 4/05/0 9/9

46 MEI Core Polynomials Section : The factor and remainder theorems Notes and Examples These notes contain subsections on The factor theorem The remainder theorem The factor theorem If x a is a factor of f(x), then f(a) = 0 and x = a is a root of the equation f(x) = 0. Conversely, if f(a) = 0, then x a is a factor of f(x). The factor theorem probably won t come as any great surprise to you. After all, you already know that you can solve some quadratics by factorising them. e.g. to solve the quadratic equation x² + x 0 = 0 you factorise: (x + 5)(x ) = 0 and deduce the solutions x = -5 and x = Clearly, for f(x) = x² + x 0, f(-5) = 0 and f() = 0. (x + 5) is a factor of f(x) f(-5) = 0 (x ) is a factor of f(x) f() = 0 The factor theorem simply extends this idea to other polynomials such as cubics, and provides a method for solving cubics and higher polynomials. Example (i) Solve the equation x³ + x² 5x 6 = 0 (ii) Sketch the graph of y = x³ + x² 5x 6 (i) The first step is to find one solution by trial and error. If there is an integer solution x = a, then by the factor theorem (x a) must be a factor of x³ + x² 5x 6. So a must be a factor of 6. a could therefore be, -,, -,, -, 6 or 6. Let f(x) = x³ + x² 5x 6 f() = = -8 f(-) = = 0 You need to find a value of x for which f(x) = 0. f(-) = 0 so by the factor theorem x + is a factor of f(x). MEI, 6/0/09 /4

47 MEI C Polynomials Section Notes and Examples The next step is to factorise f(x) into the linear factor x + and a quadratic factor. x³ + x² 5x 6 = (x + ) quadratic factor. Let the quadratic factor be ax² + bx + c. x³ + x² 5x 6 = (x + )(ax² + bx + c) = ax³ + bx² + cx + ax² + bx + c = ax³ + (a + b)x² + (b + c)x + c Multiply out the brackets Equating coefficients of x³ a = Equating constant term c = -6 Equating coefficients of x² a + b = b = Check coefficient of x: b + c = 6 = -5 x³ + x² 5x 6 = (x + )(x² + x 6) = (x + )(x )(x + ) Factorise the quadratic factor The solutions of the equation are x = -, x = and x = -. (ii) Part (i) shows that the graph of y = x³ + x² 5x 6 crosses the x-axis at (-, 0), (-, 0) and (, 0). By putting x = 0 you can see that it crosses the y-axis at (0, -6). This information allows you to sketch the graph. In the example above, the quadratic factor was found by equating coefficients. There are a number of other methods of finding this quadratic factor. You can do it by polynomial division, or by inspection (which means doing it in your head). Which one you use is really down to personal preference. The PowerPoint presentation Factorising polynomials shows these alternative methods. MEI, 6/0/09 /4

48 MEI C Polynomials Section Notes and Examples You can look at some more examples of factorising cubics using the Flash resource Factorising a cubic. You can also look at the Solving cubics video. Example f(x) = x³ + px² + 5x 6 has a factor x. Find the value of p and hence factorise f(x) as far as possible. x is a factor of f(x) f() = 0 f() = 6 + 4p = 0 + 4p 0 + 4p = 0 p = -5 f(x) = x³ - 5x² + 5x 6 x³ - 5x² + 5x 6 = (x )(ax² + bx + c) = ax³ + bx² + cx ax² - bx c = ax³ + (b a)x² + (c b)x c Equating coefficients of x³ a = Equating constant terms -c = -6 c = Equating coefficients of x² b a = -5 b 4 = -5 b = - Check coefficient of x: c b = + = 5 x³ - 5x² + 5x 6 = (x )(x² x + ) The discriminant of the quadratic factor is (-)² - 4 = 4 = - As this is negative, the quadratic factor cannot be factorised further. The remainder theorem The factor theorem is a special case of the remainder theorem. The remainder theorem says: For a polynomial f(x), f(a) is the remainder when f(x) is divided by x a. f ( x) x a g( x) f ( a) If the remainder is zero, then x a is a factor of f(x). So if x a is a factor of f(x), then f(a) = 0. This is the factor theorem. MEI, 6/0/09 /4

49 MEI C Polynomials Section Notes and Examples Don t forget about the remainder theorem! When a remainder is asked for in examination questions, there are usually quite a number of candidates who divide out to find it. This takes longer and you are much more likely to make an error. The remainder theorem is quick and straightforward to use. Example Find the remainder when f(x) = x 4 x³ + x² 4 is divided by (i) x (ii) x + (i) By the remainder theorem the remainder is f() f() = 4 ³ + ² - 4 = = -8 Remainder = -8 (ii) By the remainder theorem the remainder is f(-) f(-) = (-) 4 (-)³ + (-)² 4 = = Remainder = You can look at similar examples using the Flash resource The remainder theorem. For extra practice in examples like the one above, try the interactive questions Finding a remainder. Example 4 f(x) = x³ + ax² + bx + When f(x) is divided by x the remainder is 7 When f(x) is divided by x + the remainder is -5 Find the values of a and b. f() = 7 + a + b + = 7 a + b = 4 f(-) = -5 (-)³ + (-)²a b + = a b + = -5 9a b = 48 a b = 6 a + b = 4 a b = 6 Adding: 4a = 0 a = 5, b = - For extra practice in examples like the one above, try the interactive questions Finding a polynomial, given the remainder. MEI, 6/0/09 4/4

50 MEI Core Polynomials Section : The graphs of quadratic functions Notes and Examples These notes and examples contain subsections on The turning point of a quadratic graph Completing the square The turning point of a quadratic graph A quadratic function is usually written in the form y ax bx c, where a, b and c are constants. However, writing quadratic functions in different forms can sometimes give you additional information about the function. You have already seen that writing a quadratic function in factorised form gives you some useful information about the graph of the function. It tells you where the graph crosses the x-axis. This also applies to other polynomial functions. However, sometimes you may not be interested in where the graph cuts the axes, but you may want to know the coordinates of the maximum or minimum point of the graph (often called the vertex). There are several ways in which you can find this point. This section looks at writing quadratic equations in the form y a( x p) q (the completed square form). The first thing to notice about the completed square form is that the expression ( x p) is always positive or zero. So if a is positive, then the first term is always positive. The smallest possible value of y therefore occurs when the first term of the expression is zero, i.e. when x = p. In this case the value of y is q. The minimum point of the graph is therefore (p, q), and the line x = p is the line of symmetry of the graph. If a is negative, then the first term is always negative. The largest possible value of y therefore occurs when the first term of the expression is zero, i.e. when x = p. In this case the value of y is q. The maximum point of the graph is therefore (p, q), and the line x = p is the line of symmetry of the graph. To help you to visualise this, have a look at the interactive spreadsheet. Select either parabola, y a( x p) q, or parabola 4, y a( x p) q. Experiment with different values of p and q and notice the coordinates of the turning point in each case. MEI, 09/07/0 /6

51 MEI C Polynomials Section Notes and Examples Example For each of the following quadratic graphs, write down the equation of the line of symmetry of the graph and the coordinates of the vertex (turning point), and hence sketch the graph. (i) (ii) (iii) y x y (x ) 5 y ( x ) (i) y x Line of symmetry is x =. Minimum point is (, ). (, ) (ii) y (x ) 5 Line of symmetry is x. Minimum point is (, -5). (, -5) (iii) y ( x ) Line of symmetry is x = -. Maximum point is (-, ). (-, ) Notice that this time the vertex is a maximum point. You can look at further examples using the Flash resources Completed square form, Completed square form () and Negative completed square form. MEI, 09/07/0 /6

52 MEI C Polynomials Section Notes and Examples Example Find the equations of quadratic graphs with the given turning points. Give the equations in the form y ax bx c. (i) Minimum point (, -) (ii) Minimum point (-, ) (iii) Maximum point (4, ) (i) The equation of the graph is (ii) The equation of the graph is (iii) The equation of the graph is y ( x ) x x x x y ( x ) x x 6 9 x 6x0 y ( x 4) x 8x 6 x 8x 6 x x 8 Completing the square You have seen from Example how writing a quadratic function in the form y A( x B) C gives you important information about the graph of the function. You now need to know how to go about expressing a quadratic function in the completed square form. To start with, have another look at the interactive spreadsheet. This time, select parabola, y ax bx c and y a( x p) q, or parabola 5, y ax bx c and y a( x p) q. Leave a as for now and choose values of b and c to give a quadratic function of the form y ax bx c (shown in blue). Now look at the completed square form (shown in red). Make sure that the value of a is, and then vary the values of p and q until the two functions coincide. Compare the two equations and make sure that you can see algebraically why they represent the same function. Think about how you might find the completed square form from the form y ax bx c. Try a different quadratic function. Predict the value of p which you will need for the completed square form. When you have got the idea, try predicting the value of q as well. Then you could try varying the value of a. MEI, 09/07/0 /6

53 MEI C Polynomials Section Notes and Examples You may also find it helpful to look at the Flash resources on completing the square (one in which the coefficient of x² is, one in which the coefficient of x² is, and one in which the coefficient of x² is -). There is also a PowerPoint presentation Completing the square, which is a step-by-step demonstration of some similar examples. The examples below demonstrate the technique of completing the square. Example Write the expression x ² 4x 7 in the completed square form. Hence sketch the graph of y x² 4x 7, showing the coordinates of its turning point. First you need to find a quadratic expression which is a perfect square and which begins with x² + 4x. You do this by looking at the coefficient of x, in this case 4, and halving it. In this case you get. This tells you that the perfect square you need is (x + )². ( x )² x² 4x 4 x² 4x 7 x² 4x 4 ( x )² x is the square The + completes the square This is why the technique is called completing the square. From the completed square form, you can see that the graph has a minimum point at (-, ). To find where the graph cuts the y-axis, substitute x = 0 into the equation. There are several different approaches to writing out the working. They are all basically the same, so if you have learnt a different way which suits you, then stick to it. MEI, 09/07/0 4/6

54 MEI C Polynomials Section Notes and Examples The next example shows a situation where the coefficient of x² is not. Example 4 Write the expression x² 6x + in the completed square form. x ² 6x ( x² x ) Start by taking out the coefficient of x², in this case, as a factor. Now look at the expression inside the bracket. You need to find a quadratic expression which is a perfect square and starts with x² x. Take the coefficient of x, which is, and halve it to get perfect square you need is therefore x. ². The 9 x ² x² x 4 x² x x² x ( x ( x )² )² x² 6x [( x ( x )² )² ] In the next example, the coefficient of x² is negative, so that the graph is the other way up. This can be dealt with by taking out a factor. Example 5 Write the expression 5 + x x² in the completed square form. Hence sketch the graph of y = 5 + x x², showing the coordinates of its turning point. Start by taking out as a factor. 5 x x² ( x² x 5) Now you need a quadratic expression which is a perfect square and starts with x² x. Half the coefficient of x is, so the perfect square you need is ( x )². ( x x )² x² 4 MEI, 09/07/0 5/6

55 MEI C Polynomials Section Notes and Examples x² x 5 x² x ( x ( x 5 x x² [( x ( x 4 )² )² 4 4 )² )² ] This expression takes its maximum value when x, and this maximum value is 4. So the graph has a maximum point at,. 4 You may also like to look at the Completing the square (max/min) video. For practice in examples like the ones above, try the interactive questions Finding the turning point of a quadratic. You could also try the extension worksheet Sandcastles on Quadratic Island. MEI, 09/07/0 6/6