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Products and factors: Introduction and recap

Introduction

In this chapter you will learn how to work with algebraic expressions. You will recap some of the work on factorisation and multiplying out expressions that you learnt in earlier grades. This work will then be extended upon for Grade 10.

Recap of Earlier Work

The following should be familiar. Examples are given as reminders.

Parts of an Expression

Mathematical expressions are just like sentences and their parts have special names. You should be familiar with the following names used to describe the parts of a mathematical expression.

a · x k + b · x + c m = 0 d · y p + e · y + f 0 a · x k + b · x + c m = 0 d · y p + e · y + f 0
(1)
 Name Examples (separated by commas) term a·xka·xk ,b·xb·x, cmcm, d·ypd·yp, e·ye·y, ff  expression a·xk+b·x+cma·xk+b·x+cm, d·yp+e·y+fd·yp+e·y+f coefficient aa, bb, dd, ee exponent (or index) kk, pp base xx, yy, cc constant aa, bb, cc, dd, ee, ff variable xx, yy equation a · x k + b · x + c m = 0 a · x k + b · x + c m = 0 inequality d · y p + e · y + f ≤ 0 d · y p + e · y + f ≤ 0 binomial expression with two terms trinomial expression with three terms

Product of Two Binomials

A binomial is a mathematical expression with two terms, e.g. (ax+b)(ax+b) and (cx+d)(cx+d). If these two binomials are multiplied, the following is the result:

( a · x + b ) ( c · x + d ) = ( a x ) ( c · x + d ) + b ( c · x + d ) = ( a x ) ( c x ) + ( a x ) d + b ( c x ) + b · d = a x 2 + x ( a d + b c ) + b d ( a · x + b ) ( c · x + d ) = ( a x ) ( c · x + d ) + b ( c · x + d ) = ( a x ) ( c x ) + ( a x ) d + b ( c x ) + b · d = a x 2 + x ( a d + b c ) + b d
(2)

Exercise 1: Product of two binomials

Find the product of (3x-2)(5x+8)(3x-2)(5x+8)

Solution
1. Step 1. Multiply out and solve :
( 3 x - 2 ) ( 5 x + 8 ) = ( 3 x ) ( 5 x ) + ( 3 x ) ( 8 ) + ( - 2 ) ( 5 x ) + ( - 2 ) ( 8 ) = 15 x 2 + 24 x - 10 x - 16 = 15 x 2 + 14 x - 16 ( 3 x - 2 ) ( 5 x + 8 ) = ( 3 x ) ( 5 x ) + ( 3 x ) ( 8 ) + ( - 2 ) ( 5 x ) + ( - 2 ) ( 8 ) = 15 x 2 + 24 x - 10 x - 16 = 15 x 2 + 14 x - 16
(3)

The product of two identical binomials is known as the square of the binomial and is written as:

( a x + b ) 2 = a 2 x 2 + 2 a b x + b 2 ( a x + b ) 2 = a 2 x 2 + 2 a b x + b 2
(4)

If the two terms are ax+bax+b


and ax-bax-b

then their product is:

( a x + b ) ( a x - b ) = a 2 x 2 - b 2 ( a x + b ) ( a x - b ) = a 2 x 2 - b 2
(5)

This is known as the difference of two squares.

Factorisation

Factorisation is the opposite of expanding brackets. For example expanding brackets would require 2(x+1)2(x+1) to be written as 2x+22x+2. Factorisation would be to start with 2x+22x+2


and to end up with 2(x+1)2(x+1). In previous grades, you factorised based on common factors and on difference of squares.

Common Factors

Factorising based on common factors relies on there being common factors between your terms. For example, 2x-6x22x-6x2


can be factorised as follows:

2 x - 6 x 2 = 2 x ( 1 - 3 x ) 2 x - 6 x 2 = 2 x ( 1 - 3 x )
(6)
Investigation : Common Factors

Find the highest common factors of the following pairs of terms:

 (a) 6y;18x6y;18x (b) 12mn;8n12mn;8n (c) 3st;4su3st;4su (d) 18kl;9kp18kl;9kp (e) abc;acabc;ac (f) 2xy;4xyz2xy;4xyz  (g) 3uv;6u3uv;6u (h) 9xy;15xz9xy;15xz  (i) 24xyz;16yz24xyz;16yz  (j) 3m;45n3m;45n

Difference of Two Squares

We have seen that:

( a x + b ) ( a x - b ) = a 2 x 2 - b 2 ( a x + b ) ( a x - b ) = a 2 x 2 - b 2
(7)

Since Equation 7 is an equation, both sides are always equal. This means that an expression of the form:

a 2 x 2 - b 2 a 2 x 2 - b 2
(8)

can be factorised to

( a x + b ) ( a x - b ) ( a x + b ) ( a x - b )
(9)

Therefore,

a 2 x 2 - b 2 = ( a x + b ) ( a x - b ) a 2 x 2 - b 2 = ( a x + b ) ( a x - b )
(10)

For example, x2-16x2-16


can be written as (x2-42)(x2-42) which is a difference of two squares. Therefore, the factors of x2-16x2-16

are (x-4)(x-4) and (x+4)(x+4).

Exercise 2: Factorisation

Factorise completely: b2y5-3aby3b2y5-3aby3

Solution
1. Step 1. Find the common factors: :
b 2 y 5 - 3 a b y 3 = b y 3 ( b y 2 - 3 a ) b 2 y 5 - 3 a b y 3 = b y 3 ( b y 2 - 3 a )
(11)
Exercise 3: Factorising binomials with a common bracket

Factorise completely: 3a(a-4)-7(a-4)3a(a-4)-7(a-4)

Solution
1. Step 1. Find the common factors :
(a-4)(a-4) is the common factor
3 a ( a - 4 ) - 7 ( a - 4 ) = ( a - 4 ) ( 3 a - 7 ) 3 a ( a - 4 ) - 7 ( a - 4 ) = ( a - 4 ) ( 3 a - 7 )
(12)
Exercise 4: Factorising using a switch around in brackets

Factorise 5(a-2)-b(2-a)5(a-2)-b(2-a)

Solution
1. Step 1. Note that (2-a)=-(a-2)(2-a)=-(a-2) :
5 ( a - 2 ) - b ( 2 - a ) = 5 ( a - 2 ) - [ - b ( a - 2 ) ] = 5 ( a - 2 ) + b ( a - 2 ) = ( a - 2 ) ( 5 + b ) 5 ( a - 2 ) - b ( 2 - a ) = 5 ( a - 2 ) - [ - b ( a - 2 ) ] = 5 ( a - 2 ) + b ( a - 2 ) = ( a - 2 ) ( 5 + b )
(13)
Recap
1. Find the products of:
 (a) 2y(y+4)2y(y+4) (b) (y+5)(y+2)(y+5)(y+2) (c) (y+2)(2y+1)(y+2)(2y+1) (d) (y+8)(y+4)(y+8)(y+4) (e) (2y+9)(3y+1)(2y+9)(3y+1) (f) (3y-2)(y+6)(3y-2)(y+6)

2. Factorise:
1. 2l+2w2l+2w
2. 12x+32y12x+32y
3. 6x2+2x+10x36x2+2x+10x3
4. 2xy2+xy2z+3xy2xy2+xy2z+3xy
5. -2ab2-4a2b-2ab2-4a2b

3. Factorise completely:
 (a) 7a+47a+4 (b) 20a-1020a-10 (c) 18ab-3bc18ab-3bc (d) 12kj+18kq12kj+18kq (e) 16k2-4k16k2-4k (f) 3a2+6a-183a2+6a-18 (g) -6a-24-6a-24 (h) -2ab-8a-2ab-8a (i) 24kj-16k2j24kj-16k2j (j) -a2b-b2a-a2b-b2a (k) 12k2j+24k2j212k2j+24k2j2 (l) 72b2q-18b3q272b2q-18b3q2 (m) 4(y-3)+k(3-y)4(y-3)+k(3-y) (n) a(a-1)-5(a-1)a(a-1)-5(a-1) (o) bm(b+4)-6m(b+4)bm(b+4)-6m(b+4) (p) a2(a+7)+a(a+7)a2(a+7)+a(a+7) (q) 3b(b-4)-7(4-b)3b(b-4)-7(4-b) (r) a2b2c2-1a2b2c2-1

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Lenses

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