4.2 Factorisation (Page 2/2)

Siyavula textbooks: grade 10 Page 2 / 2

There are some tips that you can keep in mind:

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

Find the factors of $3 x^{2} + 2 x - 1$ .

Check whether the quadratic is in the form $a x^{2} + b x + c$ with $a$ positive.
The quadratic is in the required form.

Write down two brackets with an

x

in each bracket and space for the remaining terms.

(x) (x)

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

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

Option 1	Option 2
$(x - 1) (3 x + 1)$	$(x + 1) (3 x - 1)$
$3 x^{2} - 2 x - 1$	$3 x^{2} + 2 x - 1$

Check your answer
$\begin{matrix} (x + 1) (3 x - 1) & = & x (3 x - 1) + 1 (3 x - 1) \\ = & (x) (3 x) + (x) (- 1) + (1) (3 x) + (1) (- 1) \\ = & 3 x^{2} - x + 3 x - 1 \\ = & x^{2} + 2 x - 1 . \end{matrix}$
Write the final answer
The factors of $3 x^{2} + 2 x - 1$ are $(x + 1)$ and $(3 x - 1)$ .

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Factorising a trinomial

Factorise the following:

(a) $x^{2} + 8 x + 15$	(b) $x^{2} + 10 x + 24$	(c) $x^{2} + 9 x + 8$
(d) $x^{2} + 9 x + 14$	(e) $x^{2} + 15 x + 36$	(f) $x^{2} + 12 x + 36$

Factorise the following:
1. $x^{2} - 2 x - 15$
2. $x^{2} + 2 x - 3$
3. $x^{2} + 2 x - 8$
4. $x^{2} + x - 20$
5. $x^{2} - x - 20$
Find the factors for the following trinomial expressions:
1. $2 x^{2} + 11 x + 5$
2. $3 x^{2} + 19 x + 6$
3. $6 x^{2} + 7 x + 2$
4. $12 x^{2} + 8 x + 1$
5. $8 x^{2} + 6 x + 1$
Find the factors for the following trinomials:
1. $3 x^{2} + 17 x - 6$
2. $7 x^{2} - 6 x - 1$
3. $8 x^{2} - 6 x + 1$
4. $2 x^{2} - 5 x - 3$

Factorisation by grouping

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

2 x^{2} + 2 x + 3 x + 3

then we can factorise as:

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

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

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

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

Find the factors of $7 x + 14 y + b x + 2 b y$ by grouping

Determine if there are common factors to all terms
There are no factors that are common to all terms.
Determine if there are factors in common between some terms
7 is a common factor of the first two terms and $b$ is a common factor of the second two terms.
Re-write expression taking the factors into account
$7 x + 14 y + b x + 2 b y = 7 (x + 2 y) + b (x + 2 y)$
Determine if there are more common factors
$x + 2 y$ is a common factor.
Re-write expression taking the factors into account
$7 (x + 2 y) + b (x + 2 y) = (x + 2 y) (7 + b)$
Write the final answer
The factors of $7 x + 14 y + b x + 2 b y$ are $(7 + b)$ and $(x + 2 y)$ .

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Khan academy video on factorising a trinomial by grouping.

Factorisation by grouping

Factorise by grouping: $6 x + a + 2 a x + 3$
Factorise by grouping: $x^{2} - 6 x + 5 x - 30$
Factorise by grouping: $5 x + 10 y - a x - 2 a y$
Factorise by grouping: $a^{2} - 2 a - a x + 2 x$
Factorise by grouping: $5 x y - 3 y + 10 x - 6$

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Source: OpenStax, Siyavula textbooks: grade 10 maths [caps]. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11306/1.4

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