2 Solution by factorisation

It may be possible to solve a quadratic equation by factorisation using the method described for factorising quadratic expressions in HELM booklet  1.5, although you should be aware that not all quadratic equations can be easily factorised.

Example 10

Solve the equation x 2 + 5 x = 0.

Solution

Factorising and equating each factor to zero we find

x 2 + 5 x = 0 is equivalent to x ( x + 5 ) = 0

so that x = 0 and x = − 5 are the two solutions.

Example 11

Solve the quadratic equation x 2 + x − 6 = 0 .

Solution

Factorising the left hand side we find x 2 + x − 6 = ( x + 3 ) ( x − 2 ) so that

x 2 + x − 6 = 0 is equivalent to ( x + 3 ) ( x − 2 ) = 0

When the product of two quantities equals zero, at least one of the two must equal zero. In this case either ( x + 3 ) is zero or ( x − 2 ) is zero. It follows that

x + 3 = 0 ,  giving  x = − 3 or x − 2 = 0 ,  giving  x = 2

Here there are two solutions, x = − 3 and x = 2 .

These solutions can be checked quite easily by substitution back into the given equation.

Example 12

Solve the quadratic equation 2 x 2 − 7 x − 4 = 0 by factorising the left-hand side.

Solution

Factorising the left hand side: 2 x 2 − 7 x − 4 = ( 2 x + 1 ) ( x − 4 ) so 2 x 2 − 7 x − 4 = 0 is equivalent to ( 2 x + 1 ) ( x − 4 ) = 0. In this case either ( 2 x + 1 ) is zero or ( x − 4 ) is zero. It follows that   2 x + 1 = 0 , giving  x = − 1 2 or x − 4 = 0 , giving  x = 4

There are two solutions, x = − 1 2 and x = 4 .

Example 13

Solve the equation 4 x 2 + 12 x + 9 = 0 .

Solution

Factorising we find   4 x 2 + 12 x + 9 = ( 2 x + 3 ) ( 2 x + 3 ) = ( 2 x + 3 ) 2

This time the factor ( 2 x + 3 ) occurs twice. The original equation 4 x 2 + 12 x + 9 = 0 becomes

( 2 x + 3 ) 2 = 0   so that    2 x + 3 = 0

and we obtain the solution x = − 3 2 . Because the factor 2 x + 3 appears twice in the equation ( 2 x + 3 ) 2 = 0 we say that this root is a repeated solution or double root .

Task!

Solve the quadratic equation 7 x 2 − 20 x − 3 = 0 .

First factorise the left-hand side:

( 7 x + 1 ) ( x − 3 )

Equate each factor is then equated to zero to obtain the two solutions:

− 1 7 and 3

Exercises

Solve the following equations by factorisation:

1.   x 2 − 3 x + 2 = 0 2.   x 2 − x − 2 = 0 3.   x 2 + x − 2 = 0
4.   x 2 + 3 x + 2 = 0 5.   x 2 + 8 x + 7 = 0 6. x 2 − 7 x + 12 = 0
7.   x 2 − x − 20 = 0 8.   4 x 2 − 4 = 0 9.   − x 2 + 2 x − 1 = 0
10.   3 x 2 + 6 x + 3 = 0 11.   x 2 + 11 x = 0 12. 2 x 2 + 2 x = 0

The factors are found to be:

1.   1 , 2 2.   − 1 , 2 3.   − 2 , 1 4.   − 1 , − 2 5.   − 7 , − 1
6.   4 , 3 7.   − 4 , 5 8.   1 , − 1 9.  1 twice 10.   − 1 twice
11.   − 11 , 0 12.   0 , − 1