Learning outcomes EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 1 TUTORIAL 3 - FACTORISATION AND QUADRATICS On completion of this unit a learner should: 1 Know how to use algebraic methods 2 Be able to use trigonometric methods and standard formula to determine areas and volumes 3 Be able to use statistical methods to display data 4 Know how to use elementary calculus techniques. OUTCOME 1- Know how to use algebraic methods Indices and logarithms: laws of indices (a m x a n = a, = a m+n m a m n, a, (a m ) n = a mn ) n a laws of logarithms (log A + log B = log AB, log A n A = n log A, log A log B log ) B e.g. common logarithms (base 10), natural logarithms (base e), exponential growth and decay Linear equations and straight line graphs: linear equations e.g. y = mx + c; straight line graph (coordinates on a pair of labelled Cartesian axes, positive or negative gradient, intercept, plot of a straight line); experimental data e.g. Ohm s law, pair of simultaneous linear equations in two unknowns Factorisation and quadratics: multiply expressions in brackets by a number, symbol or by another expression in a bracket; by extraction of a common factor e.g. ax + ay, a(x + 2) + b(x + 2); by grouping e.g. ax - ay + bx - by, quadratic expressions e.g. a 2 + 2ab + b 2 ; roots of an equation e.g. quadratic equations with real roots by factorisation, and by the use of formula D.J.Dunn www.freestudy.co.uk 1
ROOTS AND FACTORS Consider y = A x B x C. A, B and C are the factors of y and the process is called factorisation. In the following we will only use factors that are INTEGERS meaning whole numbers. There can be many factors. For example the factors of y = 4 can be (4)(1) or (2)(2) You should know that a negative number multiplied by another negative number is a positive number but a negative number multiplied by a positive number is a negative number so we have other factors of y = 4 such as (-1)(-4) and (-2)(-2). The factors of y = -4 are (-1)(4) or (1)(-4) or (2)(-2) There is a special case when the factors are the same and also the smallest number possible, then we a root. For example the lowest factors of y = 9 are (3)(3) or (-3)(-3) The square root of 9 is 3 or -3 ( y = ±3). For example the lowest factors of y = 8 are (2)(2)(2) then the cube root of 8 is 2 y 2 3. Note that -2 is not a cube root since (-2)(-2)(-2) = -8 so it follows that -2 is the cube root of -8. Note that a positive number can have a square root that is both positive and negative. COMMON FACTORS Consider the equation y = 2A + 2B + 2C. The number 2 is common to each term so we can write the equation as y = 2(A + B + C) and you should know that the bracket indicates that every term inside is multiplied by 2. In the same way in the equation y = ax + az, a is a common factor so we can write y = a(x + z). To reverse the process all we do is multiply every term in the bracket by the factor outside the bracket. WORKED EXAMPLE No. 1 Multiply out the equation y = 3(x + 2). It should be clear that y = 3x + 6 WORKED EXAMPLE No. 2 Find the common factor of the equation y = 2x 2 + 2x - 2. It is clear that 2 appears in every term so we can write the equation as y = 2(x 2 + x - 1) 2 is a common factor. D.J.Dunn www.freestudy.co.uk 2
WORKED EXAMPLE No. 3 Simplify the equation y = ax + 2a - bz + 3b by forming brackets with common factors. Identify a is common to two terms and b is common to two terms so form two brackets with a common factor. y = a(x + 2) + b(-z + 3) or more tidily y = a(x + 2) + b(3 - z) WORKED EXAMPLE No. 4 Multiply out the brackets in the following equation and express y as a series of terms only. z = 3(y + x) - 5 (2y - x) z = 3y + 3x - 10y + 5x = 8x - 7y WORKED EXAMPLE No. 5 Find the factors of y = (3x + 6) If we write the equation as y = (3x + 3x2) we see that 3 is a common factor so: y = 3(x + 2) and the factors of y are 3 and (x-2). D.J.Dunn www.freestudy.co.uk 3
SELF ASSESSMENT EXERCISE No. 1 1. Find the square roots (x) for the following numbers. i. y = x 2 = 9 ii. y = x 2 = 16 iii. y = x 2 = 12 iv. y = x 2 = 20 2. Find the factors A, B and C of the following numbers where A, B and C are integers (whole numbers) and neither A,B nor C is 1. i. y = A B = -9 ii. y = A B = -15 iii. y = A B = -4 iv. y = A B C = 30 v. y = A B C = -30 3. Multiply out the brackets in the following equations and express y as a series of terms. i. y = 3(b - 2a) + 5(b - 2a) ii. y = 2(a - 3b) - (2b + a) iii. y = 7(a - b) + 5 (b - 2a) 4. Remove the common factors from the following equations and simplify the equations into two brackets. i. z = ax + by - ay - bx ii. z = ax - 2y + 2ax + 3by iii. z = bx 2 + cx 2 + cx + ax 5. Factorise the following into a number and a bracket. i. y = 2x + 8 ii. y = 5z 2 + 10z + 15 iii. y = 6x 2-3x + 12 D.J.Dunn www.freestudy.co.uk 4
QUADRATIC EQUATIONS and ROOTS Consider the equation y = (a + x)(b + x). The brackets are the factors of y. Consider the equation y = (a + x)(a + x). Because both brackets are the same, it follows that (a + x) must be the square root of y. We could have written y = (a + x) 2 and so y = (a + x) The point is that equations can have square roots and factors as well as numbers. DEFINITION OF A QUADRATIC EQUATION We have already dealt with proportional equations and the linear graphs produced by plotting them. In Engineering and Science, the relationship between two variables is not often proportional and often turns out to be a form of quadratic equation. Quadratic equations have the general form:- y = ax 2 + bx 1 + c x 0 Quadratic simply means that the highest power in the equation is 2 (x 2 in this case). You should know that x 1 is written as x and x 0 = 1 (anything to the power of 0 is 1) so the quadratic equation is more usually written as y = ax 2 + bx + c a, b and c are the coefficients of x 2, x 1 and x o respectively. When b = c = 0 we have the simplest quadratic y = ax 2 ROOT A special case occurs when y = 0. In this case the values of x that make y = 0 are called the roots of the equation. Suppose we have to solve y = ax 2 + bx + c If we rewrite the equation as 0 = ax 2 + bx + (c-y) and solve x, we will be solving the roots of the equation and the same values of x will give the required value of y. WORKED EXAMPLE No. 6 Rearrange the following into a standard quadratic equation ready for finding the roots. 6 = 2x 2 + 5x +2 6 = 2x 2 + 5x +2 0 = 2x 2 + 5x +2-6 0 = 2x 2 + 5x - 4 D.J.Dunn www.freestudy.co.uk 5
SOLVING WITH GRAPHS When we plot y and x for a quadratic equation, there are normally two values of x that correspond to any value of y but it is of particular interest to find the values of x when y is zero. These are called the ROOTS OF THE EQUATION. These occur where the graph crosses the x axis (i.e. where y = 0). WORKED EXAMPLE No. 7 Solve the values of x that makes z = 4 z = 2x 2-4x - 2 We are solving the values of x that makes 2x 2-4x - 2 = 4 We could simply plot z against z and get the graph shown. We see that when z = 4 that x has two values, -1 and 3. We could rearrange the equation so that 2x 2-4x - 2 4 = y y = 2x 2-4x - 6 and solve y by finding the roots which are the values at y = 0. These are also -1 and 3. It is always a good idea to check that putting these values into the equation produces y = 0. x = -1 y = 2(-1) 2-4(-1) 6 = 2 + 4 6 = 0 or z = 2x 2-4x - 2 = 2(-1) 2-4(-1) 2 = 2 + 4 2 = 4 x = 3 y = 2(3) 2-4(3) 6 = 18 12 6 = 0 or z = 2x 2-4x - 2 = 2(3) 2-4(3) 2 = 18 12 2 = 4 Note that if we plotted z and x the answers would be found at It is nice when the roots are whole numbers (integers) but this is not usually the case. Note that quadratic equations do not always cross the x axis and the solution to these requires advanced studies. D.J.Dunn www.freestudy.co.uk 6
SOLVING BY FACTORISATION A quadratic equation often factorises into two parts such that the sum of the two parts produces the original equation. In the case of equations, the factors are not numbers but expressions. Sometimes factors are easy to spot in an equation but generally it is quite difficult. If we know the factors, solving is easy as in this next example. WORKED EXAMPLE No. 8 Solve x in the (2x + 2) (x - 3) = 0 (2x + 2)(x - 3) = 0 If 2x = -2 then (-2 + 2)(x - 3) = (0)(x - 3) = 0 and so x = -1 is a solution (root) If x = 3 then (2x + 2)(3-3) = (2x + 2)(0) = 0 and so x = -3 is a solution (root) We solve the roots by simply equating each bracket to zero. Knowing how to factorise an equation takes practice. The first step in understanding this is see how multiplying two factors gives the original equation. Suppose that y = (2x + 2)(x - 3). The two factor are (2x + 2) and (x - 3). Let s multiply them. Here s how. We can multiply out the brackets as follows. y = x(2x + 2) -3(2x + 2) y = 2x 2 + 2x - 6x - 6 y = 2x 2-4x - 6 This is another way that you might do the multiplication. Factorisation is the reverse process and the bracket method is the easiest. To find the factors, this rule helps us. D.J.Dunn www.freestudy.co.uk 7
Even with this rule, you need to play around with the possibilities before you arrive at the correct answer. If we arrange that a = 1 by dividing every term by 'a' we have a quadratic equation in the form: y = 0 = x 2 + bx + c The factors will be y = 0 = (x + A)(x + B) Multiply out the brackets and y = 0 = x 2 + x(a+b) + AB Comparing coefficients we see that b = A+B and c = AB Let's not lose sight of the object which is to solve quadratic equations. So long as the numbers allow us to factorise easily, then we can solve problems like the following. WORKED EXAMPLE No. 8 Solve the values of x that satisfies the equation y = f(x) = 0 = x 2 + 5x + 6. Assume that the answers are integers. y = 0 = x 2 + 5x + 6 Assuming that the equation factorises we write: y = (x + A)(x + B) = x 2 + (A+B)x + AB Comparing coefficient we see that (A+B) = 5 and AB = 6 Try various combinations of A and B until you hit on the right one (since we are guessing the problem has been designed to give nice numbers) (AB) = 6 so this could be 1 x 6, 6 x 1, 2 x 3, 3 x 2, or The only combination that makes A + B = 5 is A = 3 and B = 2 y = 0 = (x + 3)(x + 2) Solve by equating each bracket to zero so x = -3 and -2 Check the answers x = -3 y = x 2 + 5x + 6 = (-3) 2 + (5)(-3) + 6 = 9-15 + 6 = 0 x = -2 y = x 2 + 5x + 6 = (-2) 2 + (5)(-2) + 6 = 4-10 + 6 = 0 If dividing every term by 'a' produces fractions then you might try the method in the next example. D.J.Dunn www.freestudy.co.uk 8
WORKED EXAMPLE No. 9 Solve f(x) = 0 = 2x 2 + 3x - 5 assuming that the factors only involve integers. Factorise to get two brackets. We know the first term of the two brackets must be 2x and x unless they are fractions. We must have two factors (2x + A)( x - B) = 0 We know AB must be -5 so try 1 and -5 (2x + 1)( x - 5) = 0 The middle term must be the sum of x and -10x and this gives -9x which is wrong. Try 5 and -1 (2x + 5)( x - 1) = 0 The middle term must be the sum of 5x and -2x and this gives 3x which is correct. (2x + 5)( x - 1) = 0 so the roots are x = -5/2 and x = 1 Check it in the original equation. x = -5/2 2(-5/2) 2 + 3(-5/2) - 5 = 25/2 15/2 5 = 0 correct x = 1 2(1) 2 + 3(1) - 5 = 2 + 3-5 = 0 correct SELF ASSESSMENT EXERCISE No. 2 Solve the following by factorisation. 1. 2x 2-5x 3 = 0 2. x 2 + 2x 8 = 0 3. x 2-2x 3 = 0 4. 7x 2-20x 3 = 0 PERFECT SQUARE ROOTS AND COMPLETING THE SQUARE When the two factors are the same, we have a perfect square root. Consider the case x 2 6x + 9 = 0 If we factorise x 2 6x + 9 we get (x 3) (x 3). We could write x 2 6x + 9 = (x 3) 2 = 0 and there is only one solution x = 3, and this is the perfect root. Factorisation is easier if we can change the equation so that it has a perfect square root. This is best demonstrated with an example. WORKED EXAMPLE No. 10 Solve x 2 6x + 8 = 0 If the last number was 9 we would have a perfect square root. We can do this by manipulating the equation to: x 2 6x + 9-1 = 0 or x 2 6x + 9 = 1 Factorise the left side into (x - 3)(x - 3) or (x - 3) 2 Now solve by taking the square root. (x - 3) = 1 = ±1 so x = 4 or x = 2 Check x = 4 x 2 6x + 8 = 0 4 2 6(4) + 8 = 16 24 + 8 = 0 so 4 is the root of the original equation. Check x = 2 2 2 6(2) + 8 = 4 12 + 8 = 0 so 2 is the root of the original equation. D.J.Dunn www.freestudy.co.uk 9
SELF ASSESSMENT EXERCISE No. 3 Find the roots of the following by completing the square. 1. x 2 + 6x + 5 = 0 2. x 2 + 12x + 20 = 0 3. x 2-8x + 11 = 0 4. x 2-4x - 3 = 0 QUADRATIC FORMULAE A method of finding the solution to quadratic equations that is most reliable and widely used is given here without proof. Arrange the equation into the form ax 2 + bx + c = 0 b b 2 4ac The two solutions are then given by x and this is the quadratic formula that 2a should be memorised. Most scientific/engineering calculators can find the solution direct but this formula is useful in advanced studies. WORKED EXAMPLE No. 11 Solve 2x 2 4x - 5= 0 a = 2 b = -4 c = -5 b x 2 b 4ac 2a (-4) (-4) 2(2) 4(2)(-5) 4 16 40 4 56 4 7.483 x 4 4 4 11.483 3.483 x or x 2.871or 0.871 4 4 Check the answers 2(2.871) 2 4(2.871) - 5= 0 2(-0.871) 2 4(-0.871) - 5= 0 2 SELF ASSESSMENT EXERCISE No. 4 Find the roots of the following by using the quadratic equation. 1. 3x 2 + 5x + 2 = 0 2. 5x 2-2x - 4 = 0 3. -2x 2-2x + 5 = 0 4. -5x 2 + 3x + 7 = 0 D.J.Dunn www.freestudy.co.uk 10