Mathematics. Polynomials and Quadratics. hsn.uk.net. Higher. Contents. Polynomials and Quadratics 52 HSN22100

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1 Higher Mathematics UNIT OUTCOME 1 Polnomials and Quadratics Contents Polnomials and Quadratics 5 1 Quadratics 5 The Discriminant 54 Completing the Square 55 4 Sketching Parabolas 57 5 Determining the Equation of a Parabola 59 6 Solving Quadratic Inequalities 60 7 Intersections of Lines and Parabolas 6 8 Polnomials 6 9 Snthetic Division Finding Unknown Coefficients Determining the Equation of a Graph 70 1 Iteration 7 HSN100 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Higher Mathematics Unit Mathematics OUTCOME 1 Polnomials and Quadratics 1 Quadratics A quadratic has the form provided a 0. You should alread be familiar with the following. a + b + c where a, b, and c are an numbers, The graph of a quadratic is called a parabola. There are two possible shapes: concave up (if a > 0 ) concave down (if a < 0 ) This has a minimum turning point This has a maimum turning point To find the roots of the quadratic factorisation a b c + +, we can use: b ± b the quadratic formula: = (this is not given in the eam) a EXAMPLES 1. Find the roots of. ( + 1)( ) + 1 or = 1 =. Solve ( )( ) or + 4 = 4 = 4 Page 5 HSN100

3 Higher Mathematics Unit Mathematics. Find the roots of We cannot factorise + 4 1, so we use the quadratic formula: 4 ± ( 1) = 1 4 ± = 4 ± 0 = = ± = ± 5 Note If there are two distinct solutions, the curve intersects the -ais twice. If there is one repeated solution, the turning point lies on the -ais. If b < 0 when using the quadratic formula, there are no points where the curve intersects the -ais. Page 5 HSN100

4 Higher Mathematics Unit Mathematics The Discriminant Given a + b + c, we call b the discriminant. It is the part of the quadratic formula which determines how man real roots a + b + c has. If b If b If b EXAMPLE > 0, the roots are real and unequal (distinct)., the roots are real and equal (i.e. a repeated root). < 0, the roots are not real; the do not eist. 1. Find the nature of the roots of a = 9 b = 4 c = 16 Since b Using the Discriminant b = = , the roots are real and equal. two roots one root no real roots The discriminant has man uses, usuall involving an unknown term in a quadratic. EXAMPLES. Find the values of q such that Since a = 6 b = 1 c = q 6 1 q q = has real roots, b b q q q 4q 144 q = has real roots. 0 Page 54 HSN100

5 Higher Mathematics Unit Mathematics. Show that ( k ) ( k ) ( k ) a = k + 4 b = k + c = k alwas has real roots. b ( k ) 4( k 4)( k ) 9k 1k 4 ( k 4)( 4k 8) = + + = = k 1k 4 8k = k + k + ( k 6) = + Since b ac ( k ) = + 6 0, the roots will alwas be real. Completing the Square Completing the square involves epressing = a( + p) + q. = a + b + c in the form In this form, we can determine the turning point of an parabola, including those with no real roots. The ais of smmetr is = p and the turning point is ( p, q). The following process can be used to complete the square: 1. Make sure the equation is in the form = a + b + c.. Separate the constant term ( c ) b bracketing everthing else on the RHS. Take out an constant common factor from the brackets. 4. Add on half of the coefficient squared and subtract half of the coefficient squared (note that adding on a number and taking it awa keeps the equation balanced). 5. Factorise the first three terms in the bracket. 6. State our answer in the form = a( + p) + q. EXAMPLES 1. Write = in the form = ( + p) + q. ( 6 ) 5 ( ) = + = = ( + ) 9 5 = ( + ) 14 (Step s 1-) (Step 4) (Step 5) (Step 6) Note You can alwas check our answer b epanding the brackets Page 55 HSN100

6 Higher Mathematics Unit Mathematics. Write = + 4 in the form = ( + p) + q. ( ) = + 4 ( ( ) ( ) ) ( 9 ) ( ) = = + = + 4. Write = + 8 in the form = ( + a) + b and then state: (i) the ais of smmetr, and (ii) the minimum turning point of the parabola with this equation. ( 8 ) ( ) = + = ( ) ( ) = = (i) The ais of smmetr is = 4. (ii) The minimum turning point is ( 4, 19). 4. A parabola has equation = Epress the equation in the form = ( + a) + b, and state the turning point of the parabola. ( ) ( ) = = ( ( ) ( ) ) = 4( ( 9 ) 4 ) + 7 = 4 ( ) = 4 ( ) The turning point is ( ) = , Page 56 HSN100

7 Higher Mathematics Unit Mathematics 4 Sketching Parabolas The method used to sketch the curve with equation depends on how man times the curve intersects the -ais. = a + b + c We have met curve sketching before. However, when sketching parabolas, we do not need to use calculus. We know there is onl one turning point, and we have methods for finding it. Parabolas with one or two roots Find the -ais intercepts b factorising or using the quadratic formula. Find the -ais intercept (i.e. where ). The turning point is on the ais of smmetr: O The ais of smmetr is halfwa between two distinct roots O A repeated root lies on the ais of smmetr Parabolas with no real roots There are no -ais intercepts. Find the -ais intercept (i.e. where ). Find the turning point b completing the square. EXAMPLES 1. Sketch the graph of = Since b 4 ac = ( 8 ) > 0, the parabola crosses the -ais twice. The -ais intercept ( ) : = ( 0) 8( 0) + 7 = 7 ( 0, 7 ) The -ais intercepts ( ) : ( 1)( 7) 1 or 7 = 1 = 7 ( 1, 0 ) ( 7, 0 ) The ais of smmetr lies halfwa between = 1 and = 7, i.e. = 4, so the -coordinate of the turning point is 4. Page 57 HSN100

8 Higher Mathematics Unit Mathematics We can now find the -coordinate: ( ) ( ) = = = 9 So the turning point is ( 4, 9).. Sketch the parabola with equation = ( 0) 6( 0) 9 = 9 ( 0, 9) = 6 9 Since b = ( 6) 4 ( 1) ( 9), there is a repeated root. The -ais intercept ( ) : The -ais intercept ( ) : Since there is a repeated root, (, 0) is the turning point.. Sketch the curve with equation = ( ) ( + )( + ) + = = (, 0) Since b = ( 8) 4 1< 0, there are no real roots. The -ais intercept ( ) : = ( 0) 8( 0) + 1 = 1 ( 0,1 ) So the turning point is (, 5 ). Complete the square: = O 9 = 6 9 ( ) = = ( ) O = ( ) + 5 = + = ( 4, 9) 8 1 O (, 5) Page 58 HSN100

9 Higher Mathematics Unit Mathematics 5 Determining the Equation of a Parabola In order to determine the equation of a parabola, we need two pieces of information: The -ais intercept(s) or turning point Another point on the parabola EXAMPLES 1. A parabola passes through the points ( 1, 0 ), ( 5, 0 ) and ( 0, ). Find the equation of the parabola. Since the parabola cuts the -ais where = 1 and = 5, the equation is of the form: = k ( 1)( 5) To find k, we use the point ( 0, ): = k ( 1)( 5) = k ( 0 1)( 0 5) = 5k k = 5 So the equation of the parabola is: = 5 ( 1)( 5) = 5 ( 5 + 5) = Find the equation of the parabola shown below. ( 1, 6) Since there is a repeated root, the equation is of the form: = k ( + 5) Hence = ( + ). 5 O To find k, we use ( 1, 6 ) : = k ( + 5) 6 = k ( 1+ 5) k = 6 = Page 59 HSN100

10 Higher Mathematics Unit Mathematics. Find the equation of the parabola shown below. O ( 4, ) 7 Since the turning point is ( 4, ), the equation is of the form: = a( 4) = Hence ( ) To find a, we use ( 0, 7) : a = 5 16 ( ) = a 4 ( ) 7 = a a = 5 6 Solving Quadratic Inequalities The most efficient wa of solving a quadratic inequalit is b making a rough sketch of the parabola. To do this we need to know: the shape concave up or concave down the -ais intercepts. We can then solve the quadratic inequalit b inspection of the sketch. EXAMPLES 1. Solve + 1 < 0. The parabola with equation The -ais intercepts are given b: + 1 ( + 4)( ) + 4 = 4 or = = + 1 is concave up. Make a sketch: 4 = + 1 So + 1 < 0 for 4 < <. Page 60 HSN100

11 Higher Mathematics Unit Mathematics. Find the values of for which The parabola with equation The -ais intercepts are given b: ( ) 7 6 ( + )( ) + or = = = is concave down. Make a sketch: = So for.. Solve 5 0 >. The parabola with equation The -ais intercepts are given b: 5 ( 1)( ) 1 or = 1 = = 5 is concave up. Make a sketch: = 5 1 So 5 0 > for < 1 and >. 4. Find the values of q for which ( ) Since ( ) a = 1 b = q 4 c = 1 + q q has non-real roots, b q + q q has non-real roots. ( ) ( )( ) = b 4ac q q ( )( ) q 8q 16 q = + 1 = q 4 q 4 q = + q 10q 16 < 0: We now need to solve the inequalit q 10q + 16 < 0 : The parabola with equation q q = is concave up. Page 61 HSN100

12 Higher Mathematics Unit Mathematics The -ais intercepts are given b: q 10q + 16 ( q )( q ) 8 q or q 8 Therefore b q = q = 8 < 0 for q 8 non-real roots when < q < 8. 7 Intersections of Lines and Parabolas Make a sketch: = q 10q < <, and so ( ) + q q has To determine how man times a line intersects a parabola, we substitute the equation of the line into the equation of the parabola. We can then use the discriminant or factorisation to find the number of intersections. If b > 0, the line and curve intersect twice If b, the line and curve intersect once (i.e. the line is a tangent to the curve) If b < 0, the line and the parabola do not intersect EXAMPLES 1. Show that the line = 5 is a tangent to the parabola and find the point of contact. Substitute = 5 into: + 1 ( 1)( 1) = + 5 = Since there is a repeated root, the line is a tangent at = 1. = + To find the -coordinate, substitute = 1 into the equation of the line: = 5( 1) = So the point of contact is ( 1, ). Page 6 HSN100

13 Higher Mathematics Unit Mathematics. Find the equation of the tangent to = + 1 that has gradient. The equation of the tangent is of the form = m + c, with m =, i.e. = + c. Substitute this into: c = = + c + 1 Since the line is a tangent: b ( ) 4 ( 1 c ) c 4c = 5 c = 5 4 Therefore the equation of the tangent is: = Polnomials Polnomials are epressions with one or more terms added together. Each term has a number (called the coefficient) followed b a variable (such as ) raised to a whole number power. For eample: or The degree of the polnomial is the value of its highest power, for eample: has degree has degree 18 Note that quadratics are polnomials of degree two. Also, constants are 0 polnomials of degree zero (e.g. 6 is a polnomial, since 6 = 6 ). Page 6 HSN100

14 Higher Mathematics Unit Mathematics 9 Snthetic Division Snthetic division provides a quick wa of evaluating functions defined b polnomials. For eample, consider f ( ) = Using a calculator, we find f ( 6) = 11. We can also evaluate this using snthetic division with detached coefficients. Step 1 Detach the coefficients, and write them across the top row of the table. Note that the must be in order of decreasing degree. If there is no term of a specific degree, then zero is its coefficient. Step Write the number for which ou want to evaluate the polnomial (the input number) Step Bring down the first coefficient Step 4 Multipl this b the input number, writing the result underneath the net coefficient. Step 5 Add the numbers in this column. Repeat Steps 4 and 5 until the last column has been completed. The number in the lower-right cell is the value of the polnomial for the input value, often referred to as the remainder Page 64 HSN100

15 Higher Mathematics Unit Mathematics EXAMPLE 1. Given f ( ) = + 40, evaluate f ( ) using snthetic division So f ( ). using the above process Note In this eample, the remainder is zero, so f ( ). This means + 40 when =, which means that = is a root of the equation. So + is a factor of the cubic. We can use this to help with factorisation: ( ) f ( ) = ( + ) q ( ) where q ( ) is a quadratic Is it possible to find the quadratic q ( ) using the table? Tring the numbers from the bottom row as coefficients, we find: ( + )( 0) = = = f ( ) 40 So using the numbers from the bottom row as coefficients does give the correct quadratic EXAMPLES. Show that 4 is a factor of 4 is a factor = 4 is a root Since the remainder is zero, = 4 is a root, so 4 is a factor. Page 65 HSN100

16 Higher Mathematics Unit Mathematics. Given f ( ) = (i) show that = 7 is a root of f ( ), and (ii) hence full factorise f ( ) Since the remainder is zero, = 7 is a root. Hence we have f ( ) = Show that = 5 is a root of factorise the cubic ( 7)( 7 1) = + + ( 7)( )( 4) = + Using snthetic division to factorise =, and hence full Since = 5 is a root, + 5 is a factor. ( )( ) = This does not factorise an further since b < 0. In the eamples above, we have been given a root or factor to help factorise polnomials. However, we can still use snthetic division if we do not know a factor or root. The roots of the polnomial f ( ) are numbers which divide the constant term eactl. So b tring snthetic division with these numbers, we will (eventuall) find a root. 5. Full factorise Numbers which divide 15: ± 1, ±, ± 5, ± 15. Tr = 1: ( 1) + 5( 1) 8( 1) 15 = Tr = 1: ( 1) + 5( 1) 8( 1) 15 = Note For ±1, it is simpler just to evaluate the polnomial directl, to see if this is a root Page 66 HSN100

17 Higher Mathematics Unit Mathematics Tr = : Since = is a root, is a factor. So = ( )( ) = ( )( + 1)( + 5) Using snthetic division to solve equations We can also use snthetic division to help solve equations. EXAMPLE 6. Find the roots of =. Numbers which divide 1: ± 1, ±, ±, ± 4, ± 6, ± 1. Tr = 1: ( 1) 15( 1) + 16( 1) + 1 = Tr = 1: ( 1) 15( 1) + 16( 1) + 1 Tr = : In general = For a polnomial f ( ): If f ( ) is divided b h Since = is a root, is a factor: ( )( ) 11 6 ( )( + 1)( 6) = or + 1 = 1 then the remainder is f ( ) h, and f ( h) h is a factor of f ( ). or 6 = 6 As we saw, snthetic division helps us to write f ( ) in the form + where q ( ) is called the quotient. ( h) q ( ) f ( h) Page 67 HSN100

18 Higher Mathematics Unit Mathematics EXAMPLE 7. Find the quotient and remainder when f ( ) = is divided b + 1, and epress ( ) The quotient is 4 ( ) ( )( ) f = Finding Unknown Coefficients f as ( 1) q( ) f ( h) and the remainder is, so Consider a polnomial with some unknown coefficients, such as + p p + 4, where p is a constant. If we divide the polnomial b h, then we will obtain an epression for the remainder in terms of the unknown constants. If we alread know the value of the remainder, we can solve for the unknown constants. EXAMPLES 1. Given that is a factor of is a factor = is a root. 1 1 p p p 4 + p p Since is a factor, the remainder is zero: 4 + p p = 4 p = , find the value of p. Page 68 HSN100

19 Higher Mathematics Unit Mathematics. When f ( ) = p + q q is divided b, the remainder is 6, and 1 is a factor of f ( ). Find the values of p and q. When f ( ) is divided b, the remainder is f ( ) : p q 17 4q p 4 p + q 8 p + 4q 4 p p + q 4 p + q 17 8 p + 8q 4 Since f ( ) = 6, we have: 8 p + 8q 4 = 6 8p + 8q = 40 p + q = 5 (1) Since 1 is a factor, f ( 1) : f ( 1) = p( 1) + q ( 1) 17( 1) + 4q = p + q q = p + 5q 17 i.e. p + 5q 17 () Note There is no need to use snthetic division here Solving (1) and () simultaneousl, we obtain: p + 5q = 17 () p q = 5 (1) 1 4q = 1 q = Giving p = 5 q = 5 = from (1). Hence p = and q =. Page 69 HSN100

20 Higher Mathematics Unit Mathematics 11 Determining the Equation of a Graph Given the roots, and at least one other point ling on the graph, we can establish the graph s equation. EXAMPLE 1. Find the equation of the cubic shown in the diagram below. 6 O 1 6 Step 1 Write out the roots, then rearrange to get the factors. Step The equation is then these factors multiplied together with a constant, k. Step Substitute the coordinates of a known point into this equation to find the value of k. Step 4 Replace k with this value in the equation. = 6 = = = k ( + 6)( + )( 1) Using ( 0, 6) : k ( 0 + 6)( 0 + )( 0 1) = 6 k ( )( 1)( 6) = 6 18k = 6 = ( + 6)( + )( 1) k = ( )( ) ( ) = = = Page 70 HSN100

21 Higher Mathematics Unit Mathematics Repeated Roots If a repeated root eists, then a stationar point lies on the -ais. Recall that a repeated root eists when two roots, and hence two factors, are equal. EXAMPLE. Find the equation of the cubic shown in the diagram below. 9 O = = = + So = k ( + )( ). Use ( 0, 9 ) to find k : 9 = k ( 0 + )( 0 ) 9 = k 9 k = 1 1 = = = + 9 So = ( + )( ) ( )( 6 9) ( ) 1 Page 71 HSN100

22 Higher Mathematics Unit Mathematics 1 Iteration When a root of a polnomial is not rational, an estimate of the root can be determined using an iterative process. EXAMPLE If f ( ) = 4 + 7, show that there is a real root between = 1 and =. Find this root correct to two decimal places. Evaluate f ( ) at = 1 and = : f ( 1) = 1 4( 1) ( 1) + 7 = f ( ) = 4( ) ( ) + 7 = 5 Since f ( 1) > 0 and f ( ) < 0, f ( ) has a root between = 1 and =. Tr halfwa between = 1 and = : f ( 1. 5) = ( 1. 5) ( 1. 5) + 7 = < 0 f ( 1) > 0 and f ( 1. 5) < 0 so the root is between = 1 and = Tr roughl halfwa between = 1 and = 1. 5: f ( 1. ) = 1. 4( 1. ) ( 1. ) + 7 = < 0 f ( 1. ) = 1. 4( 1. ) ( 1. ) > 0 f ( 1. ) > 0 and f ( 1. ) < 0 so the root is between = 1. and = 1.. Tr halfwa between = 1. and = 1. : f ( 1. 5) = ( 1. 5) ( 1. 5) > 0 f ( 1. 5) > 0 and f ( 1. ) < 0 so the root is between = 1. 5 and = 1.. Tr roughl halfwa between = 1. 5 and = 1. : f ( 1. 7) = ( 1. 7) ( 1. 7) > 0 f ( 1. 8) = ( 1. 8) ( 1. 8) + 7 = < 0 f ( 1. 7) > 0 and f ( 1. 8) < 0 so the root is between = 1. 7 and = Tr halfwa between = 1. 7 and = 1. 8: f ( 1.75) = ( 1. 75) ( 1. 75) f ( 1. 75) > 0 and f ( 1. 8) < 0 so the root is between = and = Therefore the root is = 1. 8 to d.p. 1 Page 7 HSN100