Products and factors (Page 4/4)

Maths test

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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 $(3 x - 1)$ .

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 $(x + 2 y)$ .

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: $5 x y - 3 y + 10 x - 6$

Simplification of fractions

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

\frac{x^{2} + 3 x}{x + 3}

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

\begin{matrix} \frac{x^{2} + 3 x}{x + 3} \\ = & \frac{x (x + 3)}{x + 3} \\ = & x provided x \neq - 3 \end{matrix}

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

Simplify: $\frac{2 x - b + x - a b}{a x^{2} - a b x}$

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

$\begin{matrix} = & \frac{(a x - a b) + (x - b)}{a x^{2} - a b x} \\ = & \frac{a (x - b) + (x - b)}{a x (x - b)} \\ = & \frac{(x - b) (a + 1)}{a x (x - b)} \end{matrix}$
Cancel out same factors
The simplified answer is:

$\begin{matrix} = & \frac{a + 1}{a x} \end{matrix}$

Simplify: $\frac{x^{2} - x - 2}{x^{2} - 4} \div \frac{x^{2} + x}{x^{2} + 2 x}$

Factorise numerators and denominators
$\begin{matrix} = & \frac{(x + 1) (x - 2)}{(x + 2) (x - 2)} \div \frac{x (x + 1)}{x (x + 2)} \end{matrix}$
Multiply by factorised reciprocal
$\begin{matrix} = & \frac{(x + 1) (x - 2)}{(x + 2) (x - 2)} \times \frac{x (x + 2)}{x (x + 1)} \end{matrix}$
Cancel out same factors
The simplified answer is

$\begin{matrix} = & 1 \end{matrix}$

Simplification of fractions

Simplify:

(a) $\frac{3 a}{15}$	(b) $\frac{2 a + 10}{4}$
(c) $\frac{5 a + 20}{a + 4}$	(d) $\frac{a^{2} - 4 a}{a - 4}$
(e) $\frac{3 a^{2} - 9 a}{2 a - 6}$	(f) $\frac{9 a + 27}{9 a + 18}$
(g) $\frac{6 a b + 2 a}{2 b}$	(h) $\frac{16 x^{2} y - 8 x y}{12 x - 6}$
(i) $\frac{4 x y p - 8 x p}{12 x y}$	(j) $\frac{3 a + 9}{14} \div \frac{7 a + 21}{a + 3}$
(k) $\frac{a^{2} - 5 a}{2 a + 10} \div \frac{3 a + 15}{4 a}$	(l) $\frac{3 x p + 4 p}{8 p} \div \frac{12 p^{2}}{3 x + 4}$
(m) $\frac{16}{2 x p + 4 x} \div \frac{6 x^{2} + 8 x}{12}$	(n) $\frac{24 a - 8}{12} \div \frac{9 a - 3}{6}$
(o) $\frac{a^{2} + 2 a}{5} \div \frac{2 a + 4}{20}$	(p) $\frac{p^{2} + p q}{7 p} \div \frac{8 p + 8 q}{21 q}$
(q) $\frac{5 a b - 15 b}{4 a - 12} \div \frac{6 b^{2}}{a + b}$	(r) $\frac{f^{2} a - f a^{2}}{f - a}$

Simplify: $\frac{x^{2} - 1}{3} \times \frac{1}{x - 1} - \frac{1}{2}$

Summary

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

End of chapter exercises

Factorise:

(a) $a^{2} - 9$	(b) $m^{2} - 36$	(c) $9 b^{2} - 81$
(d) $16 b^{6} - 25 a^{2}$	(e) $m^{2} - (1 / 9)$	(f) $5 - 5 a^{2} b^{6}$
(g) $16 b a^{4} - 81 b$	(h) $a^{2} - 10 a + 25$	(i) $16 b^{2} + 56 b + 49$
(j) $2 a^{2} - 12 a b + 18 b^{2}$	(k) $- 4 b^{2} - 144 b^{8} + 48 b^{5}$

Show that ${(2 x - 1)}^{2} - {(x - 3)}^{2}$ can be simplified to $(x + 2) (3 x - 4)$
What must be added to $x^{2} - x + 4$ to make it equal to ${(x + 2)}^{2}$

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Source: OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2

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