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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)
Table 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:

Table 2
(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:
    Table 3
    (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)
    Click here for the solution
  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
    Click here for the solution
  3. Factorise completely:
    Table 4
    (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
    Click here for the solution

More Products

Figure 1
Khan Academy video on products of polynomials.

We have seen how to multiply two binomials in "Product of Two Binomials". In this section, we learn how to multiply a binomial (expression with two terms) by a trinomial (expression with three terms). Fortunately, we use the same methods we used to multiply two binomials to multiply a binomial and a trinomial.

For example, multiply 2x+12x+1 by x2+2x+1x2+2x+1.

( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( apply distributive law ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( expand the brackets ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( group like terms to simplify ) = 2 x 3 + 5 x 2 + 4 x + 1 ( simplify to get final answer ) ( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( apply distributive law ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( expand the brackets ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( group like terms to simplify ) = 2 x 3 + 5 x 2 + 4 x + 1 ( simplify to get final answer )
(14)

Tip:

Multiplication of Binomial with Trinomial

If the binomial is A+BA+B and the trinomial is C+D+EC+D+E, then the very first step is to apply the distributive law:

( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E ) ( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E )
(15)

If you remember this, you will never go wrong!

Exercise 5: Multiplication of Binomial with Trinomial

Multiply x-1x-1 with x2-2x+1x2-2x+1.

Solution

  1. Step 1. Determine what is given and what is required :

    We are given two expressions: a binomial, x-1x-1, and a trinomial, x2-2x+1x2-2x+1. We need to multiply them together.

  2. Step 2. Determine how to approach the problem :

    Apply the distributive law and then simplify the resulting expression.

  3. Step 3. Solve the problem :
    ( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( expand the brackets ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( group like terms to simplify ) = x 3 - 3 x 2 + 3 x - 1 ( simplify to get final answer ) ( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( expand the brackets ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( group like terms to simplify ) = x 3 - 3 x 2 + 3 x - 1 ( simplify to get final answer )
    (16)
  4. Step 4. Write the final answer :

    The product of x-1x-1 and x2-2x+1x2-2x+1 is x3-3x2+3x-1x3-3x2+3x-1.

Exercise 6: Sum of Cubes

Find the product of x+yx+y
and x2-xy+y2x2-xy+y2.

Solution

  1. Step 1. Determine what is given and what is required :

    We are given two expressions: a binomial, x+yx+y, and a trinomial, x2-xy+y2x2-xy+y2.
    We need to multiply them together.

  2. Step 2. Determine how to approach the problem :

    Apply the distributive law and then simplify the resulting expression.

  3. Step 3. Solve the problem :
    ( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( expand the brackets ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( group like terms to simplify ) = x 3 + y 3 ( simplify to get final answer ) ( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( expand the brackets ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( group like terms to simplify ) = x 3 + y 3 ( simplify to get final answer )
    (17)
  4. Step 4. Write the final answer :

    The product of x+yx+y
    and x2-xy+y2x2-xy+y2
    is x3+y3x3+y3.

Tip:

We have seen that:
( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3
(18)

This is known as a sum of cubes.

Investigation : Difference of Cubes

Show that the difference of cubes (x3-y3x3-y3
) is given by the product of x-yx-y
and x2+xy+y2x2+xy+y2.

Products

  1. Find the products of:
    Table 5
    (a) (-2y2-4y+11)(5y-12)(-2y2-4y+11)(5y-12)(b) (-11y+3)(-10y2-7y-9)(-11y+3)(-10y2-7y-9)
    (c) (4y2+12y+10)(-9y2+8y+2)(4y2+12y+10)(-9y2+8y+2)(d) (7y2-6y-8)(-2y+2)(7y2-6y-8)(-2y+2)
    (e) (10y5+3)(-2y2-11y+2)(10y5+3)(-2y2-11y+2)(f) (-12y-3)(12y2-11y+3)(-12y-3)(12y2-11y+3)
    (g) (-10)(2y2+8y+3)(-10)(2y2+8y+3)(h) (2y6+3y5)(-5y-12)(2y6+3y5)(-5y-12)
    (i) (6y7-8y2+7)(-4y-3)(-6y2-7y-11)(6y7-8y2+7)(-4y-3)(-6y2-7y-11)(j) (-9y2+11y+2)(8y2+6y-7)(-9y2+11y+2)(8y2+6y-7)
    (k) (8y5+3y4+2y3)(5y+10)(12y2+6y+6)(8y5+3y4+2y3)(5y+10)(12y2+6y+6)(l) (-7y+11)(-12y+3)(-7y+11)(-12y+3)
    (m) (4y3+5y2-12y)(-12y-2)(7y2-9y+12)(4y3+5y2-12y)(-12y-2)(7y2-9y+12)(n) (7y+3)(7y2+3y+10)(7y+3)(7y2+3y+10)
    (o) (9)(8y2-2y+3)(9)(8y2-2y+3)(p) (-12y+12)(4y2-11y+11)(-12y+12)(4y2-11y+11)
    (q) (-6y4+11y2+3y)(10y+4)(4y-4)(-6y4+11y2+3y)(10y+4)(4y-4)(r) (-3y6-6y3)(11y-6)(10y-10)(-3y6-6y3)(11y-6)(10y-10)
    (s) (-11y5+11y4+11)(9y3-7y2-4y+6)(-11y5+11y4+11)(9y3-7y2-4y+6)(t) (-3y+8)(-4y3+8y2-2y+12)(-3y+8)(-4y3+8y2-2y+12)
    Click here for the solution

Factorising a Quadratic

Figure 2
Khan Academy video on factorising a quadratic.

Factorisation can be seen as the reverse of calculating the product of factors. In order to factorise a quadratic, we need to find the factors which when multiplied together equal the original quadratic.

Let us consider a quadratic that is of the form ax2+bxax2+bx
. We can see here that xx is a common factor of both terms. Therefore,
ax2+bxax2+bx
factorises to x(ax+b)x(ax+b). For example, 8y2+4y8y2+4y
factorises to
4y(2y+1)4y(2y+1).

Another type of quadratic is made up of the difference of squares. We know that:

( a + b ) ( a - b ) = a 2 - b 2 . ( a + b ) ( a - b ) = a 2 - b 2 .
(19)

This is true for any values of aa and bb, and more importantly since it is an equality, we can also write:

a 2 - b 2 = ( a + b ) ( a - b ) . a 2 - b 2 = ( a + b ) ( a - b ) .
(20)

This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down what the factors are.

Exercise 7: Difference of Squares

Find the factors of 9x2-259x2-25.

Solution

  1. Step 1. Examine the quadratic :

    We see that the quadratic is a difference of squares because:

    ( 3 x ) 2 = 9 x 2 ( 3 x ) 2 = 9 x 2
    (21)

    and

    5 2 = 25 . 5 2 = 25 .
    (22)
  2. Step 2. Write the quadratic as the difference of squares :
    9 x 2 - 25 = ( 3 x ) 2 - 5 2 9 x 2 - 25 = ( 3 x ) 2 - 5 2
    (23)
  3. Step 3. Write the factors :
    ( 3 x ) 2 - 5 2 = ( 3 x - 5 ) ( 3 x + 5 ) ( 3 x ) 2 - 5 2 = ( 3 x - 5 ) ( 3 x + 5 )
    (24)
  4. Step 4. Write the final answer :

    The factors of 9x2-259x2-25
    are (3x-5)(3x+5)(3x-5)(3x+5).

These types of quadratics are very simple to factorise. However, many quadratics do not fall into these categories and we need a more general method to factorise quadratics like x2-x-2x2-x-2
?

We can learn about how to factorise quadratics by looking at how two binomials are multiplied to get a quadratic. For example, (x+2)(x+3)(x+2)(x+3) is multiplied out as:

( x + 2 ) ( x + 3 ) = x ( x + 3 ) + 2 ( x + 3 ) = ( x ) ( x ) + 3 x + 2 x + ( 2 ) ( 3 ) = x 2 + 5 x + 6 . ( x + 2 ) ( x + 3 ) = x ( x + 3 ) + 2 ( x + 3 ) = ( x ) ( x ) + 3 x + 2 x + ( 2 ) ( 3 ) = x 2 + 5 x + 6 .
(25)

We see that the x2x2
term in the quadratic is the product of the xx-terms in each bracket. Similarly, the 6 in the quadratic is the product of the 2 and 3 in the brackets. Finally, the middle term is the sum of two terms.

So, how do we use this information to factorise the quadratic?

Let us start with factorising x2+5x+6x2+5x+6
and see if we can decide upon some general rules. Firstly, write down two brackets with an xx in each bracket and space for the remaining terms.

( x ) ( x ) ( x ) ( x )
(26)

Next, decide upon the factors of 6. Since the 6 is positive, these are:

Table 6
Factors of 6
1 6
2 3
-1 -6
-2 -3

Therefore, we have four possibilities:

Table 7
Option 1 Option 2 Option 3 Option 4
( x + 1 ) ( x + 6 ) ( x + 1 ) ( x + 6 ) ( x - 1 ) ( x - 6 ) ( x - 1 ) ( x - 6 ) ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) ( x - 2 ) ( x - 3 ) ( x - 2 ) ( x - 3 )

Next, we expand each set of brackets to see which option gives us the correct middle term.

Table 8
Option 1 Option 2 Option 3 Option 4
( x + 1 ) ( x + 6 ) ( x + 1 ) ( x + 6 ) ( x - 1 ) ( x - 6 ) ( x - 1 ) ( x - 6 ) ( x + 2 ) ( x + 3 ) ( x + 2 ) ( x + 3 ) ( x - 2 ) ( x - 3 ) ( x - 2 ) ( x - 3 )
x 2 + 7 x + 6 x 2 + 7 x + 6 x 2 - 7 x + 6 x 2 - 7 x + 6 x 2 + 5 x + 6 x 2 + 5 x + 6 x 2 - 5 x + 6 x 2 - 5 x + 6

We see that Option 3 (x+2)(x+3) is the correct solution. As you have seen that the process of factorising a quadratic is mostly trial and error, there is some information that can be used to simplify the process.

Method: Factorising a Quadratic

  1. First, divide the entire equation by any common factor of the coefficients so as to obtain an equation of the form ax2+bx+c=0ax2+bx+c=0
    where aa, bb and cc have no common factors and aa is positive.
  2. Write down two brackets with an xx in each bracket and space for the remaining terms.
    (x)(x)(x)(x)
    (27)
  3. Write down a set of factors for aa and cc.
  4. Write down a set of options for the possible factors for the quadratic using the factors of aa and cc.
  5. Expand all options to see which one gives you the correct answer.

There are some tips that you can keep in mind:

  • If cc is positive, then the factors of cc must be either both positive or both negative. The factors are both negative if bb is negative, and are both positive if bb is positive. If cc is negative, it means only one of the factors of cc is negative, the other one being positive.
  • Once you get an answer, multiply out your brackets again just to make sure it really works.

Exercise 8: Factorising a Quadratic

Find the factors of 3x2+2x-13x2+2x-1.

Solution
  1. Step 1. Check whether the quadratic is in the form ax2+bx+cax2+bx+c
    with aa positive. :

    The quadratic is in the required form.

  2. Step 2. Write down two brackets with an xx
    in each bracket and space for the remaining terms. :
    ( x ) ( x ) ( x ) ( x )
    (28)

    Write down a set of factors for aa and cc. The possible factors for aa are: (1,3). The possible factors for cc are: (-1,1) or (1,-1).

    Write down a set of options for the possible factors of the quadratic using the factors of aa and cc. Therefore, there are two possible options.

    Table 9
    Option 1 Option 2
    ( x - 1 ) ( 3 x + 1 ) ( x - 1 ) ( 3 x + 1 ) ( x + 1 ) ( 3 x - 1 ) ( x + 1 ) ( 3 x - 1 )
    3 x 2 - 2 x - 1 3 x 2 - 2 x - 1 3 x 2 + 2 x - 1 3 x 2 + 2 x - 1
  3. Step 3. Check your answer :
    ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 . ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 .
    (29)
  4. Step 4. Write the final answer :

    The factors of 3x2+2x-13x2+2x-1
    are (x+1)(x+1) and (3x-1)(3x-1).

Factorising a Trinomial

  1. Factorise the following:
    Table 10
    (a) x2+8x+15x2+8x+15
    		(b) x2+10x+24x2+10x+24
    		(c) x2+9x+8x2+9x+8
    (d) x2+9x+14x2+9x+14
    		(e) x2+15x+36x2+15x+36
    		(f) x2+12x+36x2+12x+36
    Click here for the solution
  2. Factorise the following:
    1. x2-2x-15x2-2x-15
    2. x2+2x-3x2+2x-3
    3. x2+2x-8x2+2x-8
    4. x2+x-20x2+x-20
    5. x2-x-20x2-x-20

      Click here for the solution
  3. Find the factors for the following trinomial expressions:
    1. 2x2+11x+52x2+11x+5
    2. 3x2+19x+63x2+19x+6
    3. 6x2+7x+26x2+7x+2
    4. 12x2+8x+112x2+8x+1
    5. 8x2+6x+18x2+6x+1

      Click here for the solution
  4. Find the factors for the following trinomials:
    1. 3x2+17x-63x2+17x-6
    2. 7x2-6x-17x2-6x-1
    3. 8x2-6x+18x2-6x+1
    4. 2x2-5x-32x2-5x-3

      Click here for the solution

Factorisation by Grouping

One other method of factorisation involves the use of common factors. We know that the factors of 3x+33x+3
are 3 and (x+1)(x+1). Similarly, the factors of 2x2+2x2x2+2x
are 2x2x
and (x+1)(x+1). Therefore, if we have an expression:

2 x 2 + 2 x + 3 x + 3 2 x 2 + 2 x + 3 x + 3
(30)

then we can factorise as:

2 x ( x + 1 ) + 3 ( x + 1 ) . 2 x ( x + 1 ) + 3 ( x + 1 ) .
(31)

You can see that there is another common factor: x+1x+1. Therefore, we can now write:

( x + 1 ) ( 2 x + 3 ) . ( x + 1 ) ( 2 x + 3 ) .
(32)

We get this by taking out the x+1x+1 and seeing what is left over. We have a +2x+2x
from the first term and a +3+3 from the second term. This is called factorisation by grouping.

Exercise 9: Factorisation by Grouping

Find the factors of 7x+14y+bx+2by7x+14y+bx+2by
by grouping

Solution

  1. Step 1. Determine if there are common factors to all terms :

    There are no factors that are common to all terms.

  2. Step 2. Determine if there are factors in common between some terms :

    7 is a common factor of the first two terms and bb is a common factor of the second two terms.

  3. Step 3. Re-write expression taking the factors into account :
    7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y ) 7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y )
    (33)
  4. Step 4. Determine if there are more common factors :

    x+2yx+2y
    is a common factor.

  5. Step 5. Re-write expression taking the factors into account :
    7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b ) 7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b )
    (34)
  6. Step 6. Write the final answer :

    The factors of 7x+14y+bx+2by7x+14y+bx+2by
    are (7+b)(7+b) and (x+2y)(x+2y).

Figure 3
Khan Academy video on factorising a trinomial by grouping.

Factorisation by Grouping

  1. Factorise by grouping: 6x+a+2ax+36x+a+2ax+3
    Click here for the solution
  2. Factorise by grouping: x2-6x+5x-30x2-6x+5x-30
     Click here for the solution
  3. Factorise by grouping: 5x+10y-ax-2ay5x+10y-ax-2ay
     Click here for the solution
  4. Factorise by grouping: a2-2a-ax+2xa2-2a-ax+2x
     Click here for the solution
  5. Factorise by grouping: 5xy-3y+10x-65xy-3y+10x-6
     Click here for the solution

Simplification of Fractions

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

x 2 + 3 x x + 3 x 2 + 3 x x + 3
(35)

has a quadratic in the numerator and a binomial in the denominator. You can apply the different factorisation methods to simplify the expression.

x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x provided x ≠ - 3 x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x provided x ≠ - 3
(36)

If xx were 3 then the denominator, x-3x-3, would be 0 and the fraction undefined.

Exercise 10: Simplification of Fractions

Simplify: 2x-b+x-abax2-abx2x-b+x-abax2-abx

Solution

  1. Step 1. Factorise numerator and denominator :

    Use grouping for numerator and common factor for denominator in this example.

    = ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b ) = ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b )
    (37)
  2. Step 2. Cancel out same factors :

    The simplified answer is:

    = a + 1 a x = a + 1 a x
    (38)

Exercise 11: Simplification of Fractions

Simplify:x2-x-2x2-4÷x2+xx2+2xx2-x-2x2-4÷x2+xx2+2x

Solution

  1. Step 1. Factorise numerators and denominators :
    = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 ) = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 )
    (39)
  2. Step 2. Multiply by factorised reciprocal :
    = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 ) = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 )
    (40)
  3. Step 3. Cancel out same factors :

    The simplified answer is

    = 1 = 1
    (41)

Simplification of Fractions

  1. Simplify:
    Table 11
    (a) 3a153a15
    		(b) 2a+1042a+104
    (c) 5a+20a+45a+20a+4
    		(d) a2-4aa-4a2-4aa-4
    (e) 3a2-9a2a-63a2-9a2a-6
    		(f) 9a+279a+189a+279a+18
    (g) 6ab+2a2b6ab+2a2b
    		(h) 16x2y-8xy12x-616x2y-8xy12x-6
    (i) 4xyp-8xp12xy4xyp-8xp12xy
    		(j) 3a+914÷7a+21a+33a+914÷7a+21a+3
    (k) a2-5a2a+10÷3a+154aa2-5a2a+10÷3a+154a
    		(l) 3xp+4p8p÷12p23x+43xp+4p8p÷12p23x+4
    (m) 162xp+4x÷6x2+8x12162xp+4x÷6x2+8x12
    		(n) 24a-812÷9a-3624a-812÷9a-36
    (o) a2+2a5÷2a+420a2+2a5÷2a+420
    		(p) p2+pq7p÷8p+8q21qp2+pq7p÷8p+8q21q
    (q) 5ab-15b4a-12÷6b2a+b5ab-15b4a-12÷6b2a+b
    		(r) f2a-fa2f-af2a-fa2f-a
    Click here for the solution
  2. Simplify: x2-13×1x-1-12x2-13×1x-1-12

    Click here for the solution

Summary

  • Product of two binomials
  • Factorising
  • Distributive law
  • Sum and difference of cubes
  • Factorising quadratics
  • Factorising by grouping
  • Simplifying fractions

End of Chapter Exercises

  1. Factorise:
    Table 12
    (a) a2-9a2-9(b) m2-36m2-36(c) 9b2-819b2-81
    (d) 16b6-25a216b6-25a2(e) m2-(1/9)m2-(1/9)(f) 5-5a2b65-5a2b6
    (g) 16ba4-81b16ba4-81b(h) a2-10a+25a2-10a+25(i) 16b2+56b+4916b2+56b+49
    (j) 2a2-12ab+18b22a2-12ab+18b2(k) -4b2-144b8+48b5-4b2-144b8+48b5

    Click here for the solution

  2. Show that (2x-1)2-(x-3)2(2x-1)2-(x-3)2 can be simplified to (x+2)(3x-4)(x+2)(3x-4)

    Click here for the solution
  3. What must be added to x2-x+4x2-x+4
    to make it equal to (x+2)2(x+2)2

    Click here for the solution

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