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1 Factorising quadratics An essential skill in many applications is the ability to factorise quadratic expressions. In this unit you will see that this can be thought of as reversing the process used to remove or multiply-out brackets from an expression. You will see a number of worked examples followed by a discussion of special cases which occur frequently with which you must become familiar. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that all this becomes second nature, and you eventually carry out the process of factorising simply by inspection. To help you to achieve this, the unit includes a substantial number of such exercises. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: factorise quadratic expressions understand what is meant by, and factorise, a difference of two squares understand what is meant by, and factorise, a complete square Contents 1. Introduction 2 2. Multiplying-out brackets, and quadratic expressions 2 3. Factorising quadratics 3 4. Factorising by inspection 4 5. Expressions where the coefficient of x 2 is not Special case 1 - the difference of two squares 7 7. Special case 2 - complete squares 8 8. Special case 3 - the constant term is missing 10 1 c mathcentre August 7, 2003

5 Suppose we want to factorise the quadratic expression x 2 9x 22 by inspection. We write x 2 9x 22 = ( )( ) and try to place the correct quantities in brackets. Clearly we will need an x in both terms: x 2 9x 22=(x )(x ) We want two numbers which multiply to give 22 and add to give 9. The two numbers are 11 and +2. x 2 9x 22=(x 11)(x +2) The answer should always be checked by multiplying-out the brackets again. If you can t manage to do these by inspection yet do not worry. Do it the way we did it before. Your ability to do this will improve with practice and experience. All of these examples have involved quadratic expressions where the coefficient of x 2 was 1. When the coefficient is a number other than 1 the problem is more difficult. We will look at examples like this in the next section. Exercises 2. Factorise the following. a) x 2 +8x +15 b) x 2 +10x +24 c) x 2 +9x +8 d) x 2 +15x +36 e) x 2 +2x 3 f) x 2 +2x 8 g) x 2 2x 3 h) x 2 +7x 18 i) x 2 12x +35 j) x 2 11x +10 k) x 2 13x +22 l) x 2 +12x Expressions where the coefficient of x 2 is not 1. Suppose we wish to factorise the expression 3x 2 +5x 2. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. However, instead of looking for two numbers which multiply to give 2 we must look for two numbers which multiply to give 6, (that is, the coefficient of x 2 multiplied by the constant term, 3 2). This is entirely consistent with the method we applied before because in those examples the coefficient of x 2 was always 1. So??= 6? +? = 5 By inspection, or trial and error, two such numbers are 6 and = =5 We use these two numbers to split the 5x term into 6x and 1x. 3x 2 +5x 2 = 3x 2 +6x x 2 = 3x(x +2) x 2 by factorising the first two terms = 3x(x +2) (x + 2) factorising the last two terms by extracting 1 = (x + 2)(3x 1) by noting the common factor of x +2 5 c mathcentre August 7, 2003

6 Suppose we wish to factorise 2x 2 +5x 7. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. We look for two numbers which multiply to give 14, (that is, the coefficient of x 2 multiplied by the constant term, 2 7). So??= 14? +?=5 By inspection, or trial and error, two such numbers are 7 and = =5 We use these two numbers to split the 5x term into 7x and 2x. 2x 2 +5x 7 = 2x 2 +7x 2x 7 = x(2x +7) 2x 7 by factorising the first two terms = x(2x +7) (2x + 7) factorising the last two terms by extracting 1 = (2x + 7)(x 1) by noting the common factor of 2x +7 Suppose we wish to factorise 6x 2 5x 4. We look for two numbers which multiply to give 24, (that is, the coefficient of x 2 multiplied by the constant term, 6 4). We look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. So??= 24? +? = 5 By inspection, or trial and error, two such numbers are 8 and = = 5 We use these two numbers to split the 5x term into 3x and 8x. 6x 2 5x 4 = 6x 2 +3x 8x 4 = 3x(2x +1) 8x 4 by factorising the first two terms = 3x(2x +1) 4(2x + 1) factorising the last two terms by extracting 4 = (2x + 1)(3x 4) by noting the common factor of 2x +1 Suppose we wish to factorise 15x 2 3x 18. In this example note that the coefficients share a common factor of 3. This can be extracted: Now we are looking at factorising 5x 2 x 6. 3(5x 2 x 6) c mathcentre August 7,

7 We look for two numbers which multiply to give 30, (that is, the coefficient of x 2 multiplied by the constant term, 5 6). We look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 1. So??= 30? +? = 1 By inspection, or trial and error, two such numbers are 5 and = = 1 We use these two numbers to split the x term into 5x and 6x. 3(5x 2 x 6) = 3(5x 2 +5x 6x 6) = 3(5x(x +1) 6x 6)) by factorising the first two terms = 3(5x(x +1) 6(x + 1)) factorising the last two terms by extracting 6 = 3((x + 1)(5x 6)) by noting the common factor of x +1 Exercises 3 Factorise the following. a) 2x 2 +11x + 5 b) 3x 2 +19x + 6 c) 3x 2 +17x 6 d) 6x 2 +7x +2 e) 7x 2 6x 1 f) 12x 2 +7x + 1 g) 8x 2 +6x + 1 h) 8x 2 6x +1 6.Specialcase1-thedifferenceoftwosquares A special case concerns what is known as the difference of two squares. What does this look like? Typically, x 2 9. Note that there is no x term and that the number 9 is itself a square number. A square number is one which has resulted from squaring another number. In this case 9 is the result of squaring 3, (3 2 = 9), and so 9 is a square number. Hence x 2 9 is the difference of two squares, x When we try to factorise x 2 9 we are looking for two numbers which add to zero (because there is no term in x), and which multiply to give 9. Two such numbers are 3 and 3 because 3 3 = 9, and = 0 We use these two numbers to split the 0x term into 3x and 3x. x 2 9 = x 2 3x +3x 9 = x(x 3)+3x 9 by factorising the first two terms = x(x 3)+3(x 3) factorising the last two terms by extracting +3 = (x 3)(x + 3) by noting the common factor of x 3 So x 2 9=(x 3)(x +3) Recall that 9 is a square number, (9 = 3 2 ). So this example is a difference of two square. The result of this example is true in more general cases of the difference of two squares. It is always the case that x 2 a 2 factorises to (x a)(x + a). 7 c mathcentre August 7, 2003

8 Key Point The difference of two squares, x 2 a 2, always factorises to x 2 a 2 =(x a)(x + a) Suppose we want to factorise x Note that x 2 25 is the difference of two squares because 25 is a square number (25 = 5 2 ). So we need to factorise x We can do this by inspection using the formula above. x =(x 5)(x +5) Suppose we want to factorise 2x This time there is a common factor of 2 which can be extracted first. 2x 2 32=2(x 2 16)=2(x 4)(x +4) A slightly different form occurs if we now include a square number in front of the x 2 term. For example, suppose we wish to factorise 9x Note that 9 is a square number, and so the term 9x 2 can be written (3x) 2. So we still have a difference of two squares (3x) To factorise this we write 9x 2 16 = (3x 4)(3x +4) Exercises 4 Factorise the following. a) x b) x 2 36 c) x 2 1 d) x e) 9x 2 1 f) 16x 2 9 g) 49x 2 1 h) 25x Specialcase2-complete squares A second important special case is when we need to deal with complete squares. This happens when the answer can be written in the form ( ) 2, that is as a single term, squared. Consider the following example. c mathcentre August 7,

9 Suppose we wish to factorise x 2 +10x As before we want two numbers which multiply to give 25 and add to give 10. Two such numbers are 5 and 5 because 5 5 = 25, and = 10 We use these two numbers to split the 10x term into 5x and 5x. x 2 +10x +25 = x 2 +5x +5x +25 = x(x +5)+5x + 25 by factorising the first two terms = x(x +5)+5(x + 5) factorising the last two terms by extracting +5 = (x + 5)(x + 5) by noting the common factor of x +5 The result can be written as (x +5) 2,acomplete square. Suppose we want to factorise x 2 8x +16. We seek two numbers which add to give 8 and multiply to give 16. The two required numbers are 4 and 4 and so, by inspection, x 2 8x +16=(x 4)(x 4) The result can be written as (x 4) 2,acomplete square. More complicated examples can occur, for example when there is a number in front of the x 2. Work through the following example. Suppose we want to factorise 25x 2 20x +4. Note that 25x 2 can be written as (5x) 2, a squared term. Note also that 4 = 2 2. In this case, by inspection, 25x 2 20x +4=(5x 2)(5x 2) The result can be written as (5x 2) 2,acomplete square. Do not worry if you have difficulty with this last example. The skill will come with practice. Even if you did not recognise the complete square, you could still use the previous method: To factorise 25x 2 20x +4 As before we want two numbers which multiply to give 100 (the coefficient of x 2 multiplied by the constant term) and add to give 20. Two such numbers are 10 and 10 because = 100, and = 20 We use these two numbers to split the 20x term into 10x and 10x. 25x 2 20x +4 = 25x 2 10x 10x +4 = 5x(5x 2) 10x + 4 by factorising the first two terms = 5x(5x 2) 2(5x 2) factorising the last two terms by extracting 2 = (5x 2)(5x 2) by noting the common factor of 5x 2 So 25x 2 20x +4=(5x 2) 2 as before. 9 c mathcentre August 7, 2003

10 8.Specialcase3-theconstanttermismissing There is one more special case that you ought to be familiar with. This arises in examples where the constant term is absent, as in 3x 2 8x. How do we proceed? The way forward is to look for common factors. In the case of 3x 2 8x note that each of the two terms has a common factor of x. (Note that this is not the case when dealing with a quadratic expression containing a constant term). The common factor, x, is written outside a bracket, and the contents of the bracket are chosen by inspection so that they multiply-out to give the correct terms. 3x 2 8x = x(3x 8) Suppose we wish to factorise 10x 2 +5x. This time we note the common factor of 5x. This is extracted and we adjust the contents of the brackets accordingly: 10x 2 +5x =5x(2x +1) Exercises 5. Factorise the following. a) x 2 +18x +81 b) x 2 4x +4 c) x 2 22x d) 25x 2 +40x +16 e) 64x 2 +16x +1 f) x 2 +6x g) 3x 2 x h) 5x 2 20x Answers Exercises 1 a) x 2 +3x +2 b) x 2 +5x +6 c) x 2 2x 3 d) x 2 2x 8 e) x 2 +4x 21 f) x 2 7x g) x 2 5x +6 h) x 2 14x +49 i) x 2 4x +4 j) x 2 4 k) x 2 9 l) 2x 2 +3x +1 m) 2x 2 11x + 5 n) 9x 2 1 o) 4x 2 8x. Exercises 2 a) (x + 3)(x +5) b) (x + 4)(x +6) c) (x + 1)(x +8) d) (x + 3)(x + 12) e) (x + 3)(x 1) f) (x + 4)(x 2) g) (x 3)(x +1) h) (x + 9)(x 2) i) (x 7)(x 5) j) (x 1)(x 10) k) (x 11)(x 2) l) (x + 9)(x +3) Exercises 3 a) (2x + 1)(x +5) b) (3x + 1)(x +6) c) (3x 1)(x +6) d) (2x + 1)(3x +2) e) (7x + 1)(x 1) f) (3x + 1)(4x +1) g) (2x + 1)(4x +1) h) (2x 1)(4x 1) Exercises 4 a) (x 12)(x + 12) b) (x 6)(x +6) c) (x 1)(x +1) d) (x 11)(x + 11) e) (3x + 1)(3x 1) f) (4x + 3)(4x 3) g) (7x + 1)(7x 1) h) (5x + 4)(5x 4) Exercises 5 a) (x +9) 2 b) (x 2) 2 c) (x 11) 2 d) (5x +4) 2 e) (8x +1) 2 f) x(x +6) g) x(3x 1) h) 5x(x 4) c mathcentre August 7,

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