Factorising expressions

Factorise the following expressions.

a) $15x+25y$

b) $2x+4xy$

c) $7pq+2ab-21bq+4ap$

Solution:

a)

$15x+25y$

In this case, there is no need to rearrange the expression to find the GCD. What is the greatest number or expression that can divide both 15x and 25y without a remainder? The GCD is 5.

Next, divide each of the expressions by 5 to get the result (macronym). Thus,

$$\begin{gathered} \frac{(15x+25y)}{5}=\frac{15x}{5}+\frac{25y}{5} \\ \frac{(15x+25y)}{5}=3x+5y \end{gathered}$$

Now, open a bracket, place the GCD outside the bracket and the result (macronym) inside the bracket. You should arrive at,

$\mathbf{5}\mathbf{(}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{)}$

This means that when factorised:

$\mathbf{15}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{5}\mathbf{(}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{)}$

b)

$2x+4xy$

In this case, there is no need to rearrange the expression to find the GCD. What is the greatest number or expression that can divide both 2x and 4xy without a remainder? The GCD is 2x.

Next, divide each of the expressions by 2x to get the result (macronym). Thus,

$$\begin{gathered} \frac{(2x+4xy)}{2x}=\frac{2x}{2x}+\frac{4xy}{2x} \\ \frac{(2x+4xy)}{2x}=1+2y \end{gathered}$$

Now, open a bracket, place the GCD outside the bracket and the result (macronym) inside the bracket. You should arrive at,

$\mathbf{2}\mathit{x}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{)}$

This implies that when factorised:

$\mathbf{2}\mathit{x}\mathbf{+}\mathbf{4}\mathit{x}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{)}$

c)

$7pq+2ab-21bq+4ap$

The first thing to do is to rearrange the expression by bringing like terms. Use the best possible route to achieve this. For instance, 2ab and -21bq can be grouped but only b is the GCD between terms in that pair compared with when 2ab and 4ap are grouped which can be done as below,

$7pq+2ab-21bq+4ap=7pq-21bq+2ab+4ap$

The next thing to do is to separate pairs with brackets to easily see what groups you are factorising, make sure you have a plus sign between brackets.

$7pq-21bq+2ab+4ap=(7pq-21bq)+(2ab+4ap)$

Now factorise what you have in the bracket separately by following previously explained steps. So,

$$\begin{gathered} \frac{(7pq-21bq)}{7q}=\frac{7pq}{7q}-\frac{21bq}{7q} \\ \frac{(7pq-21bq)}{7q}=p-3b \\ 7pq-21bq=7q(p-3b) \end{gathered}$$

Similarly, for the other pair, we have

$$\begin{gathered} \frac{(2ab+4ap)}{2a}=\frac{2ab}{2a}+\frac{4ap}{2a} \\ \frac{(2ab+4ap)}{2a}=b+2p \\ 2ab+4ap=2a(b+2p) \end{gathered}$$

Now you have successfully factorised pairs, bring them back together but this time in their factorized forms. Thus,

$\mathbf{7}\mathit{q}\mathbf{(}\mathit{p}\mathbf{-}\mathbf{3}\mathit{b}\mathbf{)}\mathbf{+}\mathbf{2}\mathit{a}\mathbf{(}\mathit{b}\mathbf{+}\mathbf{2}\mathit{p}\mathbf{)}$

This implies that:

$\mathbf{7}\mathit{p}\mathit{q}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{a}\mathit{b}\mathbf{-}\mathbf{21}\mathit{b}\mathit{q}\mathbf{+}\mathbf{4}\mathit{a}\mathit{p}\mathbf{=}\mathbf{7}\mathit{q}\mathbf{(}\mathit{p}\mathbf{-}\mathbf{3}\mathit{b}\mathbf{)}\mathbf{+}\mathbf{2}\mathit{a}\mathbf{(}\mathit{b}\mathbf{+}\mathbf{2}\mathit{p}\mathbf{)}$

Factorise the following.

a) $4x+2y-10$

b) $3x+y-9$

Solution:

a)

$4x+2y-10$

Find the GCD, the GCD here is 2, divide by 2 using the same steps explained in previous examples and you would get,

$4x+2y-10=\mathbf{2}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\mathit{y}\mathbf{-}\mathbf{5}\mathbf{)}$

b)

$3x+y-9$

Here, there is no common factor among all three terms, however, there is a common factor among two of the terms. This means that you can group them together to arrive at,

$3x+y-9=3x-9+y$

Now, put a bracket separating the terms you are to factorise with a common factor from that which cannot be divided by the common factor.

$3x-9+y=(3x-9)+y$

Now you can factorize what is in the bracket by the GCD, which is 3. Avoid tampering with the other term(s) outside the bracket.

$$\begin{gathered} \frac{(3x-9)}{3}=\frac{3x}{3}-\frac{9}{3} \\ \frac{3x}{3}-\frac{9}{3}=x-3 \end{gathered}$$

So you have factorised to get,

$3x-9=3(x-3)$

Now you can add back the other term(s) outside the bracket to complete the expression.

$3x+y-9=\mathbf{3}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{+}\mathit{y}$

Factorise the following expressions.

a) $x^2-5x+6$

b) $2x^2+9x+4$

Solution:

a)

$x^2-5x+6$

The first thing to do is to multiply the constant 6 by x² to get 6x². Recall that the product rule says,

$\alpha \beta =ac$

ac is the coefficient of x² and in this case it is 6.

Now, write out the factor product of 6 considering both positive and negative numbers. This includes:

$2\times 3=6$,

$-2\times (-3)=6$,

$1\times 6=6$

and

$-1\times (-6)=6$

Now you have possible factors, you need to know which pair would comply with the sum rule. Recall that:

$\alpha +\beta =b$

So we are looking for the sum of the factors that would give the coefficient of x which is -5 from this question. Let's look at the sum of the pairs.

$$\begin{gathered} 2+3=5 \\ \mathbf{-}\mathbf{2}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{5} \\ 1+6=7 \\ -1+(-6)=-7 \end{gathered}$$

This means that our right factor pair is -2 and -3.

The next thing is to replace -5x with our factors knowing that,

$$\begin{gathered} -5x=-2x+(-3x) \\ -5x=-2x-3x \end{gathered}$$

Therefore,

$x^2-5x+6=x^2-2x-3x+6$

Now just follow the earlier explained steps and factorise by first placing your brackets and having a plus sign to separate both brackets.

$x^2-2x-3x+6=(x^2-2x)+(-3x+6)$

Then factorise with the GCD of each expression in the brackets.

$(x^2-2x)+(-3x+6)=x(x-2)-3(x-2)$

In factorising quadratic equations, you must ensure, that the results (macronyms) are the same in both brackets. In this case, we have (x-2) in both brackets.

The next step is quite interesting, add the factors outside the bracket and put them inside a bracket, and eliminate one of the similar brackets so that you have two brackets separated with no sign between the brackets. Do this and you would have:

$\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}$

Therefore,

$x^2-5x+6=\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}$

b)

$2x^2+9x+4$

The first thing to do is to multiply the constant 4 by 2x² to get 8x². Recall that the product rule says,

$\alpha \beta =ac$

ac is the coefficient of x² and in this case, it is 8.

Now, write out the factor product of 8 considering both positive and negative numbers. This includes:

$2\times 4=8$,

$-2\times (-4)=8$,

$1\times 8=8$

and

$-1\times (-8)=8$

Now you have possible factors, you need to know which pair would comply with the sum rule. Recall that:

$\alpha +\beta =b$

So we are looking for the sum of the factors that would give the coefficient of x which is 9 from this question. Let's look at the sum of the pairs.

$$\begin{gathered} 2+4=6 \\ -2+(-4)=-6 \\ \mathbf{1}\mathbf{+}\mathbf{8}\mathbf{=}\mathbf{9} \\ -1+(-8)=-9 \end{gathered}$$

This means that our right factor pair is 1 and 8.

The next thing is to replace 9x with our factors knowing that,

$9x=x+8x$

Therefore,

$2x^2+9x+4=2x^2+x+8x+4$

Now just follow the earlier explained steps and factorise by first placing your brackets and having a plus sign to separate both brackets.

$2x^2+x+8x+4=(2x^2+x)+(8x+4)$

Then factorise with the GCD of each expression in the brackets.

$(2x^2+x)+(8x+4)=x(2x+1)+4(2x+1)$

In factorising quadratic equations, you must ensure, that the results (macronyms) are the same in both brackets. In this case, we have (2x+1) in both brackets.

The next step is quite interesting, add the factors outside the bracket and put them inside a bracket, and eliminate one of the similar brackets so that you have two brackets separated with no sign between the brackets. Do this and you would have:

$\mathbf{(}\mathit{x}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}$

Therefore,

$2x^2+9x+4=\mathbf{(}\mathit{x}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}$

Solve the following.

$14{(k+1)}^2+21(k+1)$

Solution:

To factorise:

$14{(k+1)}^2+21(k+1)$

you have to expand the expression. Thus,

$$\begin{gathered} 14{(k+1)}^2+21(k+1)=14(k+1)(k+1)+21k+21 \\ (k+1)(k+1)=k^2+2k+1 \\ 14(k+1)(k+1)+21k+21=14(k^2+2k+1)+21k+21 \\ 14(k^2+2k+1)+21k+21=14k^2+28k+14+21k+21 \end{gathered}$$

Bring like terms together.

$$\begin{gathered} 14k^2+28k+14+21k+21=14k^2+28k+21k+14+21 \\ 14k^2+28k+21k+14+21=14k^2+49k+35 \end{gathered}$$

Looking at the expression$14k^2+49k+35$, you can tell that a factor is common which is 7. Apply the steps explained earlier and arrive at:

$14k^2+49k+35=7(2k^2+7k+5)$

Now, the expression$7(2k^2+7k+5)$can be further factorised by factorising $(2k^2+7k+5)$which is a quadratic expression. Try out what you have learnt so far on factorising the expression $2k^2+7k+5$ and you should arrive at:

$2k^2+7k+5=(2k+5)(k+1)$

Do not forget you have a 7 multiplying through the expression, so the factorised expression is,

$$\begin{gathered} 7(2k^2+7k+5)=7(2k+5)(k+1) \\ 14{(k+1)}^2+21(k+1)=7(2k+5)(k+1) \end{gathered}$$

Finicky was given 7 oranges and 4 pears while Indodo was 3 oranges and 11 pears. Express the sum of their fruits in a factorised expression.

Solution:

Let the oranges be x and pears be y. Thus Finicky's fruits received can be expressed as,

$7x+4y$

while Indodo's fruits received can be expressed as,

$3x+11y$

The sum of Finicky's and Indodo's fruits would be:

$$\begin{gathered} (7x+4y)+(3x+11y)=7x+3x+4y+11y \\ 7x+3x+4y+11y=10x+15y \end{gathered}$$

Now you have the sum, factorise the expression.

$10x+15y=\mathbf{5}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{)}$