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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). We can 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

Adding and subtracting fractions

Using the concepts learnt in simplification of fractions, we can now add and subtract simple fractions. To add or subtract fractions we note that we can only add or subtract fractions that have the same denominator. So we must first make all the denominators the same and then perform the addition or subtraction. This is called finding the lowest common denominator or multiple.

For example, if you wanted to add: 1212 and 3535 we would note that the lowest common denominator is 10. So we must multiply the first fraction by 5 and the second fraction by 2 to get both of these with the same denominator. Doing so gives: 510510 and 610610. Now we can add the fractions. Doing so, we get 11101110.

Exercise 12

Simplify the following expression: x-2x2-4+x2x-2-x3+x-4x2-4x-2x2-4+x2x-2-x3+x-4x2-4

Solution

  1. Step 1. Factorise numerators and denominators :
    x-2 (x+2)(x-2) + x2 x-2 - x3+x-4 (x+2)(x-2) x-2 (x+2)(x-2) + x2 x-2 - x3+x-4 (x+2)(x-2)
    (42)
  2. Step 2. Make the denominators the same :

    We make all the denominators the same so that we can add or subtract the fractions. The lowest common denominator is (x-2)(x+2)(x-2)(x+2).

    x-2 (x+2)(x-2) + (x2) (x+2) (x+2)(x-2) - x3+x-4 (x+2)(x-2) x-2 (x+2)(x-2) + (x2) (x+2) (x+2)(x-2) - x3+x-4 (x+2)(x-2)
    (43)
  3. Step 3. Write all the fractions as one:

    Since the fractions all have the same denominator we can write them all as one fraction with the appropriate operator

    x-2 + (x2) (x+2) - x3+x-4 (x+2)(x-2) x-2 + (x2) (x+2) - x3+x-4 (x+2)(x-2)
    (44)
  4. Step 4. Simplify the numerator:
    2x2 +2x-6 (x+2)(x-2) 2x2 +2x-6 (x+2)(x-2)
    (45)
  5. Step 5. Write the final answer:
    2(x2 +x-3) (x+2)(x-2) 2(x2 +x-3) (x+2)(x-2)
    (46)

Two interesting mathematical proofs

We can use the concepts learnt in this chapter to demonstrate two interesting mathematical proofs. The first proof states that n2+nn2+n is even for all nZnZ. The second proof states that n3-nn3-n is divisible by 6 for all nZnZ. Before we demonstrate that these two laws are true, we first need to note some other mathematical rules.

If we multiply an even number by an odd number, we get an even number. Similarly if we multiply an odd number by an even number we get an even number. Also, an even number multiplied by an even number is even and an odd number multiplied by an odd number is odd. This result is shown in the following table:

Table 12
  Odd number Even number
Odd number Odd Even
Even number Even Even

If we take three consecutive numbers and multiply them together, the resulting number is always divisible by three. This should be obvious since if we have any three consecutive numbers, one of them will be divisible by 3.

Now we are ready to demonstrate that n2+nn2+n is even for all nZnZ. If we factorise this expression we get: n(n+1)n(n+1). If nn is even, than n+1n+1 is odd. If nn is odd, than n+1n+1 is even. Since we know that if we multiply an even number with an odd number or an odd number with an even number, we get an even number, we have demonstrated that n2+nn2+n is always even. Try this for a few values of nn and you should find that this is true.

To demonstrate that n3-nn3-n is divisible by 6 for all nZnZ, we first note that the factors of 6 are 3 and 2. So if we show that n3-nn3-n is divisible by both 3 and 2, then we have shown that it is also divisible by 6! If we factorise this expression we get: n(n+1)(n-1)n(n+1)(n-1). Now we note that we are multiplying three consecutive numbers together (we are taking nn and then adding 1 or subtracting 1. This gives us the two numbers on either side of nn.) For example, if n=4n=4, then n+1=5n+1=5 and n-1=3n-1=3. But we know that when we multiply three consecutive numbers together, the resulting number is always divisible by 3. So we have demonstrated that n3-nn3-n is always divisible by 3. To demonstrate that it is also divisible by 2, we can also show that it is even. We have shown that n2+nn2+n is always even. So now we recall what we said about multiplying even and odd numbers. Since one number is always even and the other can be either even or odd, the result of multiplying these numbers together is always even. And so we have demonstrated that n3-nn3-n is divisible by 6 for all nZnZ.

Summary

  • A binomial is a mathematical expression with two terms. The product of two identical binomials is known as the square of the binomial. The difference of two squares is when we multiply ( a x + b ) ( a x - b ) ( a x + b ) ( a x - b )
  • Factorising is the opposite of expanding the brackets. You can use common factors or the difference of two squares to help you factorise expressions.
  • 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 ) ) helps us to multiply a binomial and a trinomial.
  • The sum of cubes is: ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 and the difference of cubes is: x3-y3=(x-y)(x2+xy+y2)x3-y3=(x-y)(x2+xy+y2)
  • To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.
  • We can also factorise a quadratic by grouping. This is where we find a common factor in the quadratic and take it out and then see what is left over.
  • We can simplify fractions by using the methods we have learnt to factorise expressions.
  • Fractions can be added or subtracted. To do this the denominators of each fraction must be the same.

End of Chapter Exercises

  1. Factorise:
    1. a2-9a2-9
    2. m2-36m2-36
    3. 9b2-819b2-81
    4. 16b6-25a216b6-25a2
    5. m2-(1/9)m2-(1/9)
    6. 5-5a2b65-5a2b6
    7. 16ba4-81b16ba4-81b
    8. a2-10a+25a2-10a+25
    9. 16b2+56b+4916b2+56b+49
    10. 2a2-12ab+18b22a2-12ab+18b2
    11. -4b2-144b8+48b5-4b2-144b8+48b5
    Click here for the solution
  2. Factorise completely:
    1. (16x4)(16x4)
    2. 7x214x+7xy14y7x214x+7xy14y
    3. y27y30y27y30
    4. 1xx2+x31xx2+x3
    5. 3(1p2)+p+13(1p2)+p+1
    Click here for the solution
  3. Simplify the following:
    1. (a-2)2-a(a+4)(a-2)2-a(a+4)
    2. (5a-4b)(25a2+20ab+16b2)(5a-4b)(25a2+20ab+16b2)
    3. (2m-3)(4m2+9)(2m+3)(2m-3)(4m2+9)(2m+3)
    4. (a+2b-c)(a+2b+c)(a+2b-c)(a+2b+c)
    Click here for the solution
  4. Simplify the following:
    1. p2-q2p÷p+qp2-pqp2-q2p÷p+qp2-pq
    2. 2x+x2-2x32x+x2-2x3
    Click here for the solution
  5. 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
  6. 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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