Factorising quadratics - Mathematics resources

1 Factorising quadraticsAn essential skill in many applications is the ability to factorise quadratic expressions. Inthis unit you will see that this can be thought of as reversing the process used to remove or multiply-out brackets from an will see a number of worked examples followed by a discussion of special cases which occurfrequently with which you must become order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that all this becomes second nature, and you eventually carry out the process offactorising simply by inspection. To help you to achieve this, the unit includes a substantialnumber of such 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 brackets, and quadratic by where the coefficient ofx2is not case 1 - the difference of two case 2 - complete case 3 - the constant term is missing101c mathcentreAugust7,20031.

2 IntroductionIn this unit you will learn how many quadratic expressions can be factorised. Essentially, thisis the reverse process of removing brackets from expressions such as (x+ 2)(x+ 3).2. Multiplying-out brackets, and quadratic expressionsYou will be familiar already with the well-known process of multiplying-out brackets. Forexample, you will have seen expressions like (x+ 2)(x+ 3) and then expanded these terms toarrive at a quadratic expression. Let us see how this works. The arrows in the figure belowshow how each term in the first pair of brackets multiplies each term in the second pair.(x+2)(x+3)=x2+3x+2x+6=x2+5x+6 Then the like-terms, 3xand 2x, are collected together to give 5x. The result of multiplying-outthe brackets is thequadratic expressionx2+5x+ PointAquadratic expressionhas the general formax2+bx+cwherea,bandcare that in a quadratic expression the highest power ofxis 2.

3 The numberais called thecoefficientofx2,bis called thecoefficientofx, andcis called theconstant term. Thesenumbers can be positive or negative. The numbersbandccan also be examples of quadratic expressions:x2 7x+11,4x2+3x 1,x2+8x,3x2+2 Like many processes in Mathematics , it is useful to be able to go the other way. That is, startingwith the quadratic expressionx2+5x+6, can we carry out a process which will result in the form(x+ 2)(x+ 3)? The answer is: yes we can! This process is calledfactorising the quadraticexpression. This would help, for example, if we wanted to solve a quadratic equation. Suchan equation is formed when we set a quadratic expression equal to zero, as inx2+5x+6=0 This is because the equation can then be written(x+ 2)(x+3)=0and if we have two expressions multiplied together resulting in zero, then one or both of thesemust be zero.

4 So, eitherx+2=0,orx+ 3 = 0, from which we can conclude thatx= 2, orx= 3. We have found the solutions of the quadratic equationx2+5x+6= mathcentreAugust7,20032So, the ability to factorise a quadratic is a useful basic skill, which you will learn about in thisunit. However, be warned, not all quadratics will factorise, but a lot do and so this is a processyou have got to get to know!Exercises 1 Prepare yourself for Factorising quadratic expressions by multiplying out the brackets in each ofthe following cases:a) (x+ 1)(x+2) b) (x+ 2)(x+3) c) (x+ 1)(x 3)d) (x+ 2)(x 4) e) (x 3)(x+7) f)x(x 7)g) (x 3)(x 2) h) (x 7)(x 7) i) (x 2)(x 2)j) (x+ 2)(x 2) k) (x 3)(x+3) l) (2x+ 1)(x+1)m) (x 5)(2x 1) n) (3x 1)(3x+1) o) 4x(x 2)3. Factorising quadraticsTo learn how to factorise let us study again the removal of brackets from (x+ 3)(x+2).

5 (x+ 3)(x+2)=x2+2x+3x+6=x2+5x+6 Clearly the number 6 in the final answer comes frommultiplyingthe numbers 3 and 2 in thebrackets. This is an important observation. The term 5xcomes fromaddingthe terms , if we were to begin withx2+5x+ 6 and we were going to reverse the process we need tolook for two numbers which multiply to give 6 and add to give ?=6?+?=5 What are these numbers ? Well, we know that they are 3 and 2, and you will learn with practiseto find these simply by the two numbers which add to give 5 we split the 5xterm into 3xand 2x. We can setthe calculation out as +5x+6 =x2+3x+2x+6=x(x+3)+2x+ 6 by Factorising the first two terms=x(x+3)+2(x+ 3) by Factorising the last two terms=(x+ 3)(x+ 2)by noting the common factor ofx+3 The quadratic has been factorised. Note that you should never get these wrong, because theanswer can always be checked by multiplying-out the brackets again!

6 ExampleSuppose we want to factorise the quadratic expressionx2 7x+ as before we look for two numbers which multiply together to give 12 and add togetherto give 7. Think about this for a minute and you will realise that the two numbers we seekare 3 and 4 because 4 3=12,and 4+ 3= 7So, using the two numbers which add to give 7 we split the 7xterm into 4xand the calculation out like this:3c mathcentreAugust7,2003x2 7x+12 =x2 4x 3x+12=x(x 4) 3x+ 12 by Factorising the first two terms=x(x 4) 3(x 4) by Factorising the last two terms extractinga factor of 3 in order to leavex 4=(x 4)(x 3)by noting the common factor ofx 4 Once again, note that the answer can be checked by multiplying out the brackets we wish to factorise the quadratic expressionx2 5x as before we look for two numbers which multiply together to give 14 and add togetherto give 5.

7 Think about this for a minute and you will realise that the two numbers we seekare 7 and 2 because 7 2= 14,and 7+2= 5So, using the two numbers which add to give 5 we split the 5xterm into 7xand + the calculation out like this:x2 5x 14 =x2 7x+2x 14=x(x 7)+2x 14 by Factorising the first two terms=x(x 7)+2(x 7) by Factorising the last two terms=(x 7)(x+ 2)by noting the common factor ofx 7So the factorisation ofx2 5x 14 is (x 7)(x+2).4. Factorising by inspectionClearly, when you have some experience of this method, it is possible to avoid writing out allthese stages. This is rather long-winded. What you need to do now is carry out the processby inspection. That is, simply by looking at the given expression, decide in your head whichnumbers need to be placed in the brackets to do the job. Look at the following two examplesand see if you can do we want to factorise the quadratic expressionx2 9x+ 20 by writex2 9x+20=()()and try to place the correct quantities in brackets.

8 Clearly we will need anxin both terms:x2 9x+20=(x)(x)We want two numbers which multiply to give 20 and add to give 9. With practice you will beable to do this in your head. The two numbers are 4 and 9x+20=(x 4)(x 5)The answer should always be checked by multiplying-out the brackets mathcentreAugust7,20034 ExampleSuppose we want to factorise the quadratic expressionx2 9x 22 by writex2 9x 22 = ( )( )and try to place the correct quantities in brackets. Clearly we will need anxin both terms:x2 9x 22=(x)(x)We want two numbers which multiply to give 22 and add to give 9. The two numbers are 11 and + 9x 22=(x 11)(x+2)The answer should always be checked by multiplying-out the brackets you can t manage to do these by inspection yet do not worry. Do it the way we did it ability to do this will improve with practice and of these examples have involved quadratic expressions where the coefficient ofx2was the coefficient is a number other than 1 the problem is more difficult.

9 We will look atexamples like this in the next the )x2+8x+15 b)x2+10x+24 c)x2+9x+8 d)x2+15x+36e)x2+2x 3f)x2+2x 8g)x2 2x 3h)x2+7x 18i)x2 12x+35 j)x2 11x+10 k)x2 13x+22 l)x2+12x+275. Expressions where the coefficient ofx2is not we wish to factorise the expression 3x2+5x we did before we look for two numbers which add to give the coefficient ofx; so we seek twonumbers which add to give , instead of looking for two numbers which multiply to give 2 we must look for twonumbers which multiply to give 6, (that is, the coefficient ofx2multiplied by the constantterm, 3 2). This is entirely consistent with the method we applied before because in thoseexamples the coefficient ofx2was always ?= 6?+?=5By inspection, or trial and error, two such numbers are 6 and 1= 66+ 1=5We use these two numbers to split the 5xterm into 6xand +5x 2=3x2+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 ofx+25c mathcentreAugust7,2003 ExampleSuppose we wish to factorise 2x2+5x we did before we look for two numbers which add to give the coefficient ofx; so we seek twonumbers which add to give look for two numbers which multiply to give 14, (that is, the coefficient ofx2multipliedby the constant term, 2 7).

10 So? ?= 14? + ?=5By inspection, or trial and error, two such numbers are 7 and 2= 147 + 2=5We use these two numbers to split the 5xterm into 7xand +5x 7=2x2+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 ExampleSuppose we wish to factorise 6x2 5x look for two numbers which multiply to give 24, (that is, the coefficient ofx2multipliedby the constant term, 6 4).We look for two numbers which add to give the coefficient ofx; so we seek two numbers whichadd to give ?= 24? + ? = 5By inspection, or trial and error, two such numbers are 8 and 8= 243 + 8= 5We use these two numbers to split the 5xterm into 3xand 5x 4=6x2+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 ExampleSuppose we wish to factorise 15x2 3x this example note that the coefficients share a common factor of 3.