Factorisation

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Introduction

In grade 10, the basics of solving linear equations, quadratic equations, exponential equations and linear inequalities were studied. This chapter extends on that work. We look at different methods of solving quadratic equations.

Solution by factorisation

How to solve quadratic equations by factorisation was discussed in Grade 10. Here is an example to remind you of what is involved.

Solve the equation $2 x^{2} - 5 x - 12 = 0$ .

Determine whether the equation has common factors
This equation has no common factors.
Determine if the equation is in the form $a x^{2} + b x + c$ with $a > 0$
The equation is in the required form, with $a = 2$ , $b = - 5$ and $c = - 12$ .
Factorise the quadratic
$2 x^{2} - 5 x - 12$ has factors of the form:

$(2 x + s) (x + v)$

with $s$ and $v$ constants to be determined. This multiplies out to

$2 x^{2} + (s + 2 v) x + s v$

We see that $s v = - 12$ and $s + 2 v = - 5$ . This is a set of simultaneous equations in $s$ and $v$ , but it is easy to solve numerically. All the options for $s$ and $v$ are considered below.

$s$ $v$ $s + 2 v$

2 -6 -10

-2 6 10

3 -4 -5

-3 4 5

4 -3 -2

-4 3 2

6 -2 2

-6 2 -2

We see that the combination $s = 3$ and $v = - 4$ gives $s + 2 v = - 5$ .
Write the equation with factors
$(2 x + 3) (x - 4) = 0$
Solve the equation
If two brackets are multiplied together and give 0, then one of the brackets must be 0, therefore

$2 x + 3 = 0$

or

$x - 4 = 0$

Therefore, $x = - \frac{3}{2}$ or $x = 4$
Write the final answer
The solutions to $2 x^{2} - 5 x - 12 = 0$ are $x = - \frac{3}{2}$ or $x = 4$ .

It is important to remember that a quadratic equation has to be in the form $a x^{2} + b x + c = 0$ before one can solve it using the following methods.

Solve for $a$ : $a (a - 3) = 10$

Rewrite the equation in the form $a x^{2} + b x + c = 0$
Remove the brackets and move all terms to one side.

$a^{2} - 3 a - 10 = 0$
Factorise the trinomial
$(a + 2) (a - 5) = 0$
Solve the equation
$a + 2 = 0$

or

$a - 5 = 0$

Solve the two linear equations and check the solutions in the original equation.
Write the final answer
Therefore, $a = - 2$ or $a = 5$

Solve for $b$ : $\frac{3 b}{b + 2} + 1 = \frac{4}{b + 1}$

Put both sides over the LCM
$\frac{3 b (b + 1) + (b + 2) (b + 1)}{(b + 2) (b + 1)} = \frac{4 (b + 2)}{(b + 2) (b + 1)}$
Determine the restrictions
The denominators are the same, therefore the numerators must be the same.

However, $b \neq - 2$ and $b \neq - 1$
Simplify equation to the standard form
$\begin{matrix} 3 b^{2} + 3 b + b^{2} + 3 b + 2 & = & 4 b + 8 \\ 4 b^{2} + 2 b - 6 & = & 0 \\ 2 b^{2} + b - 3 & = & 0 \end{matrix}$
Factorise the trinomial and solve the equation
$\begin{matrix} (2 b + 3) (b - 1) & = & 0 \\ 2 b + 3 = 0 & o r & b - 1 = 0 \\ b = \frac{- 3}{2} & o r & b = 1 \end{matrix}$
Check solutions in original equation
Both solutions are valid

Therefore, $b = \frac{- 3}{2}$ or $b = 1$

Solution by factorisation

Solve the following quadratic equations by factorisation. Some answers may be left in surd form.

$2 y^{2} - 61 = 101$
$2 y^{2} - 10 = 0$
$y^{2} - 4 = 10$
$2 y^{2} - 8 = 28$
$7 y^{2} = 28$
$y^{2} + 28 = 100$
$7 y^{2} + 14 y = 0$
$12 y^{2} + 24 y + 12 = 0$
$16 y^{2} - 400 = 0$
$y^{2} - 5 y + 6 = 0$
$y^{2} + 5 y - 36 = 0$
$y^{2} + 2 y = 8$
$- y^{2} - 11 y - 24 = 0$
$13 y - 42 = y^{2}$
$y^{2} + 9 y + 14 = 0$
$y^{2} - 5 k y + 4 k^{2} = 0$
$y (2 y + 1) = 15$
$\frac{5 y}{y - 2} + \frac{3}{y} + 2 = \frac{- 6}{y^{2} - 2 y}$
$\frac{y - 2}{y + 1} = \frac{2 y + 1}{y - 7}$

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Source: OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3

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$s$	$v$	$s + 2 v$
2	-6	-10
-2	6	10
3	-4	-5
-3	4	5
4	-3	-2
-4	3	2
6	-2	2
-6	2	-2