1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style

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1 Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with in this Block. It is essential that you have had plenty practice in removing brackets before you attempt this Block. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... identify common factors in an expression 1 be able to remove brackets Learning Style To achieve what is expected of you... allocate sufficient study time factorise simple expressions factorise quadratic expressions briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. Factorisation A number is said to be factorised when it is written as a product. For example 21 can be factorised into 7 3. We say that 7 and 3 are factors of 21. Always remember that the factors of a number are multiplied together. Algebraic expressions can also be factorised. Consider the expression 7(2x + 1). Removing the brackets we can rewrite this as 7(2x + 1) = 7(2x) + (7)(1) = 14x +7. Thus 14x + 7 is equivalent to 7(2x + 1). We see that 14x + 7 has factors 7 and (2x + 1). The factors 7 and (2x +1)multiply together to give 14x + 7. The process of writing an expression as a product of its factors is called factorisation. When asked to factorise 14x + 7 we write 14x + 7 = 7(2x +1) and so we see that, here, factorisation can be regarded as reversing the process of removing brackets. Always remember that the factors of an algebraic expression are multiplied together. Example factorise the expression 4x Solution Both terms in the expression 4x + 20 are examined to see if they have any factors in common. Clearly 20 can be factorised as (4)(5) and so we can write 4x +20=4x + (4)(5) The factor 4 is common to both terms on the right; it is called a common factor and is placed at the front and outside brackets to give 4x +20=4(x +5) Note that the factorised form can and should be checked by removing the brackets again. Example factorise z 2 5z. Solution Note that since z 2 = z z we can write z 2 5z = z(z) 5z so that there is a common factor of z. Hence z 2 5z = z(z) 5z = z(z 5) Engineering Mathematics: Open Learning Unit Level 0 2

3 Example factorise 6x 9y. Solution By observation we note that there is a common factor of 3. Thus 6x 9y = 3(2x) 3(3y) = 3(2x 3y) Try each part of this exercise Identify the factor common to both 14z and 21w. Hence factorise 14z +21w. Part (a) First find the factor common to both 14z and 21w Part (b) Now complete the problem Try each part of this exercise factorise 6x 12xy. Part (a) First identify any common factors. In this case there are two Part (b) Now complete the problem. If there is any doubt, check your answer by removing the brackets again. More exercises for you to try 1. Factorise a) 5x +15y, b)3x 9y, c)2x +12y, d)4x +32z +16y, e) 1x + 1y. 2 4 In each case check your answer by removing the brackets again. 2. Factorise a) a 2 +3ab, b)xy + xyz, c)9x 2 12x 3. Explain why a is a factor of a + ab but b is not. Factorise a + ab. 4. Explain why x 2 is a factor of 4x 2 +3yx 3 +5yx 4 but y is not. Factorise 4x 2 +3yx 3 +5yx 4. 3 Engineering Mathematics: Open Learning Unit Level 0

4 2. factorising quadratic expressions One of the most important forms, occurring in many areas of mathematics, physics and engineering, is the quadratic expression. Many quadratic expressions can be written as the product of two linear factors and, in this section, we examine how these can be easily found. Key Point An expression of the form ax 2 + bx + c where a, b and c are numbers is called a quadratic expression. The numbers b and c may be zero but a must not be zero (for, then, the quadratic reduces to a linear expression). The number a is called the coefficient of x 2, b is the coefficient of x and c is called the constant term. Consider the product (x + 1)(x + 2). Removing brackets yields x 2 +3x + 2. Conversely, we see that the factors of x 2 +3x + 2 are (x + 1) and (x + 2). However, if we were given the quadratic expression first, how would we factorise it? The following examples show how to do this but note that not all quadratic expressions can be easily factorised. To enable us to factorise a quadratic expression in which the coefficient of x 2 equals 1, we note the following expansion: (x + m)(x + n) =x 2 + mx + nx + mn = x 2 +(m + n)x + mn So, given a quadratic expression we can think of the coefficient of x as m + n and the constant term as mn. Once the values of m and n have been found the factors can be easily obtained. Example Factorise x 2 +4x 5. Solution Writing x 2 +4x 5=(x + m)(x + n) =x 2 +(m + n)x + mn we seek numbers m and n so that m + n = 4 and mn = 5. By trial and error it is not difficult to find that m = 5 and n = 1 (or, the other way round, m = 1 and n = 5). So we can write x 2 +4x 5=(x + 5)(x 1) The answer can be checked easily by removing brackets. Try each part of this exercise Factorise x 2 +6x +8. The coefficient of x 2 is 1. We can write x 2 +6x +8=(x + m)(x + n) =x 2 +(m + n)x + mn so that m + n = 6 and mn =8. Engineering Mathematics: Open Learning Unit Level 0 4

5 Part (a) Try various possibilities for m and n until you find values which satisfy both of these equations. Part (b) Finally factorise the quadratic: When the coefficient of x 2 is not equal to 1 it may be possible to extract a numerical factor. For example, note that 3x 2 +18x + 24 can be written as 3(x 2 +6x + 8) and then factorised as in the previous example. Sometimes no numerical factor can be found and a slightly different approach may be taken. We will demonstrate a technique which can always be used to transform the given expression into one in which the coefficient of the squared variable equals 1. Example Factorise 2x 2 +5x +3. Solution First note the coefficient of x 2 ; in this case 2. Multiply the whole expression by this number and rearrange as follows: 2(2x 2 +5x + 3) = 2(2x 2 ) + 2(5x) + 2(3) = (2x) 2 + 5(2x)+6. If we now introduce a new variable such that z =2x we find that the coefficient of the squared term equals 1. Thus we can write (2x) 2 + 5(2x)+6 as z 2 +5z +6 This can be factorised to give (z +3)(z +2). Returning to the original variable by writing z =2x we find 2(2x 2 +5x +3)=(2x + 3)(2x +2) A factor of 2 can be extracted from the second bracket on the right so that 2(2x 2 +5x + 3) = 2(2x + 3)(x +1) so that 2x 2 +5x +3=(2x + 3)(x +1) As an alternative to the technique of the previous example, experience and practice can often help us to identify factors. For example suppose we wish to factorise 3x 2 +7x + 2. We write 3x 2 +7x +2=( )( ) In order to obtain the term 3x 2 we can place terms 3x and x in the brackets to give 3x 2 +7x +2=(3x +? )(x +? ) In order to obtain the constant 2, we consider the factors of 2. These are 1,2 or 1, 2. By placing these factors in the brackets we can factorise the quadratic expression. Various possibilities exist: 5 Engineering Mathematics: Open Learning Unit Level 0

6 we could write (3x + 2)(x + 1), (3x + 1)(x + 2), (3x 2)(x 1) or (3x 1)(x 2), only one of which is correct. By removing brackets from each in turn we look for the factorisation which produces the correct middle term, 7x. The correct factorisation is found to be 3x 2 +7x +2=(3x + 1)(x +2) With practice you will be able to carry out this process quite easily. Try each part of this exercise factorise the quadratic expression 5x 2 7x 6 Write 5x 2 7x 6=( )( ) To obtain the quadratic term 5x 2, insert 5x and x in the brackets: Part (a) Now examine the factors of 6. 5x 2 7x 6=(5x +?)(x +?) Part (b) Use these factors to find which pair, if any, gives rise to the middle term, 7x, and complete the factorisation. On occasions you will meet expressions of the form x 2 y 2. Such an expression is known as the difference of two squares. Note that here we are finding the difference between two squared terms. It is easy to verify by removing brackets that this factorises as x 2 y 2 =(x + y)(x y) So, if you can learn to recognise such expressions it is an easy matter to factorise them. Example Factorise a) x 2 36z 2, b) 25x 2 9z 2, c) α 2 1 Solution In each case we are required to find the difference of two squared terms. (a) Note that x 2 36z 2 = x 2 (6z) 2. This factorises as (x +6z)(x 6z). (b) Here 25x 2 9z 2 =(5x) 2 (3z) 2. This factorises as (5x +3z)(5x 3z). (c) α 2 1=(α + 1)(α 1). Engineering Mathematics: Open Learning Unit Level 0 6

7 More exercises for you to try 1. Factorise a) x 2 +8x +7,b)x 2 +6x 7, c) x 2 +7x + 10, d) x 2 6x +9,e)x 2 +5x Factorise a) 2x 2 +3x +1,b)2x 2 +4x +2,c)3x 2 3x 6, d) 5x 2 4x 1, e) 16x 2 1, f) x 2 +1,g) 2x 2 + x Factorise a) x 2 +9x + 14, b) x 2 +11x + 18, c) x 2 +7x 18, d) x 2 +4x 77, e) x 2 +2x, f) 3x 2 + x, g)3x 2 +4x +1,h)6x 2 +5x +1,i)6x 2 +31x + 35, j) 6x 2 +7x 5, k) 3x 2 +2x +5,l)x 2 3x Factorise a) z 2 144, b) z 2 1 4,c)s Engineering Mathematics: Open Learning Unit Level 0

8 End of Block 1.5 Engineering Mathematics: Open Learning Unit Level 0 8

9 7 9 Engineering Mathematics: Open Learning Unit Level 0

10 7(2z +3w) Engineering Mathematics: Open Learning Unit Level 0 10

11 6 and x 11 Engineering Mathematics: Open Learning Unit Level 0

12 6x(1 2y) Engineering Mathematics: Open Learning Unit Level 0 12

13 1. a) 5(x +3y), b) 3(x 3y), c) 2(x +6y), d) 4(x +8z +4y), e) 1 2 (x y) 2. a) a(a +3b), b) xy(1 + z), c) 3x(3x 4). 3. a(1 + b). 4. x 2 (4+3yx +5yx 2 ). 13 Engineering Mathematics: Open Learning Unit Level 0

14 e.g. m =4,n= 2 or, the other way round, m =2,n=4 Engineering Mathematics: Open Learning Unit Level 0 14

15 (x + 4)(x +2) 15 Engineering Mathematics: Open Learning Unit Level 0

16 3, 2 or 3,2 or 6,1 or 6, 1 Engineering Mathematics: Open Learning Unit Level 0 16

17 (5x + 3)(x 2) 17 Engineering Mathematics: Open Learning Unit Level 0

18 1. a) (x + 7)(x + 1), b) (x + 7)(x 1), c) (x + 2)(x + 5), d) (x 3)(x 3), e) (x + 3)(x +2) 2. a) (2x + 1)(x + 1), b) 2(x +1) 2,c)3(x + 1)(x 2), d)(5x + 1)(x 1), e) (4x + 1)(4x 1), f) (x + 1)(1 x), g) (x + 1)(3 2x) 3. a) (7 + x)(2 + x), b) (9 + x)(2 + x), c) (x + 9)(x 2), d) (x + 11)(x 7), e) (x +2)x, f) (3x +1)x, g) (3x + 1)(x + 1), h) (3x + 1)(2x +1) i) (3x + 5)(2x + 7), j) (3x + 5)(2x 1), k) (5 3x)(x + 1), l) (2 x)(1 x) 4. a) (z + 12)(z 12), b) (z + 1)(z 1), c) (s + 1)(s 1) Engineering Mathematics: Open Learning Unit Level 0 18

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