# Factorising expressions

On a well celebrated university event called "rag day"  I and my friend Biodun dressed up in different costumes. However, people who saw us said we were brothers because although we had different costumes on, they were both martial arts costumes. Herein, you shall be learning about factorising expressions, a concept that was used subconsciously by our observers.

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## What is factorising expressions?

Factorising expressions is the process of simplifying expressions which gives rise to the greatest common divisor (GCD) outside the bracket and the result inside the bracket.

An expansion between the GCD and the result would reproduce the original expression.

Recall at the beginning of this study, I and my friend Biodun had dressed in costumes and were thought to be related. We were actually factorized because observers saw what was similar between us.

In expressions such as those discussed here, similarities are determined by finding the greatest common factor. For instance, the numbers 6 and 4 have a number common among both. This number must be able to divide both 6 and 4 without a remainder, that number is 2.

Similarly, 12 and 8 have 2 as their divisor, however, 4 is also a divisor. In this case, we choose 4 because it is a greater number than 2 and would reduce 8 and 12 to the smallest results.

When 2 divides 12 and 8 the result is 6 and 4 respectively. Meanwhile, 4 would divide 12 and 8 into 3 and 2 respectively.

## How is factorising of simple expressions calculated?

In order to factorise simple expressions, you must follow the following steps:

1. Rearrange expression when necessary - this is done by bringing like terms.

2. Find the greatest common divisor (GCD) among terms in the expression.

3. Divide terms by the GCD.

4. Open a bracket.

5. Place the GCD outside the bracket and the result of the division inside the bracket.

Therefore, your factorised expression should be in the form:

$a\left(x±y\right)$

Where,

a is the GCD

and $x±y$ is the result.

I often refer to what is inside the brackets as a macronym meaning math-acronym because most acronyms are found inside a bracket

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,

$\frac{\left(15x+25y\right)}{5}=\frac{15x}{5}+\frac{25y}{5}\phantom{\rule{0ex}{0ex}}\frac{\left(15x+25y\right)}{5}=3x+5y$

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

$\mathbf{5}\mathbf{\left(}\mathbf{3}\mathbit{x}\mathbf{+}\mathbf{5}\mathbit{y}\mathbf{\right)}$

This means that when factorised:

$\mathbf{15}\mathbit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{25}\mathbit{y}\mathbf{=}\mathbf{5}\mathbf{\left(}\mathbf{3}\mathbit{x}\mathbf{+}\mathbf{5}\mathbit{y}\mathbf{\right)}$

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,

$\frac{\left(2x+4xy\right)}{2x}=\frac{2x}{2x}+\frac{4xy}{2x}\phantom{\rule{0ex}{0ex}}\frac{\left(2x+4xy\right)}{2x}=1+2y$

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

$\mathbf{2}\mathbit{x}\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbit{y}\mathbf{\right)}$

This implies that when factorised:

$\mathbf{2}\mathbit{x}\mathbf{+}\mathbf{4}\mathbit{x}\mathbit{y}\mathbf{=}\mathbf{2}\mathbit{x}\mathbf{\left(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbit{y}\mathbf{\right)}$

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.

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

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

$\frac{\left(7pq-21bq\right)}{7q}=\frac{7pq}{7q}-\frac{21bq}{7q}\phantom{\rule{0ex}{0ex}}\frac{\left(7pq-21bq\right)}{7q}=p-3b\phantom{\rule{0ex}{0ex}}7pq-21bq=7q\left(p-3b\right)$

Similarly, for the other pair, we have

$\frac{\left(2ab+4ap\right)}{2a}=\frac{2ab}{2a}+\frac{4ap}{2a}\phantom{\rule{0ex}{0ex}}\frac{\left(2ab+4ap\right)}{2a}=b+2p\phantom{\rule{0ex}{0ex}}2ab+4ap=2a\left(b+2p\right)$

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

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

This implies that:

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

## How do you factorise and expand linear expressions?

Linear expressions are algebraic representations where both constants and variables are all in the first power. They are in the form:

$3x+y-6$

however, the expression,

$3x+{y}^{2}-6$

is not a linear expression because y is in the second power.

In factorising linear expressions, the earlier outlined steps still apply.

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,

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

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.

$3x-9+y=\left(3x-9\right)+y$

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.

$\frac{\left(3x-9\right)}{3}=\frac{3x}{3}-\frac{9}{3}\phantom{\rule{0ex}{0ex}}\frac{3x}{3}-\frac{9}{3}=x-3$

So you have factorised to get,

$3x-9=3\left(x-3\right)$

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

$3x+y-9=\mathbf{3}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{3}\mathbf{\right)}\mathbf{+}\mathbit{y}$

Note that when the factorized expression is expanded would give the original expression. Should you attempt expanding and fail to arrive at the original expression, then, that is an indication that you must have gotten one or more steps wrong. Hence, you are advised to refactorise the expression till you get it right.

## How do you factorise quadratic expressions?

Quadratic expressions are expressions in the form:

$a{x}^{2}+bx+c$

where a, b and c are numbers not equal to zero.

You factorise quadratic expressions by using the sum and product rule.

### Sum and product rule of factorising quadratic expressions

The sum rule of quadratic expressions states that the sum of the two factors of the quadratic equation is equal to the coefficient of bx in the equation,

$a{x}^{2}+bx+c$

which is b. If the factors of the quadratic equation are α and β, then:

$\alpha +\beta =b$

Meanwhile, the product rule states that when those two factors multiply themselves, the product is equal to the coefficient of acx2. Thus,

$\alpha \beta =ac$

Therefore, when finding the factors of a quadratic equation you must ensure that your found factors comply with the sum and product rule.

Factorise the following expressions.

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

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

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:

$2×3=6$,

$-2×\left(-3\right)=6$,

$1×6=6$

and

$-1×\left(-6\right)=6$

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.

$2+3=5\phantom{\rule{0ex}{0ex}}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{\left(}\mathbf{-}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{5}\phantom{\rule{0ex}{0ex}}1+6=7\phantom{\rule{0ex}{0ex}}-1+\left(-6\right)=-7$

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

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

$-5x=-2x+\left(-3x\right)\phantom{\rule{0ex}{0ex}}-5x=-2x-3x$

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.

${x}^{2}-2x-3x+6=\left({x}^{2}-2x\right)+\left(-3x+6\right)$

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

$\left({x}^{2}-2x\right)+\left(-3x+6\right)=x\left(x-2\right)-3\left(x-2\right)$

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:

$\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{3}\mathbf{\right)}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{2}\mathbf{\right)}$

Therefore,

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

b)

$2{x}^{2}+9x+4$

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:

$2×4=8$,

$-2×\left(-4\right)=8$,

$1×8=8$

and

$-1×\left(-8\right)=8$

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.

$2+4=6\phantom{\rule{0ex}{0ex}}-2+\left(-4\right)=-6\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{+}\mathbf{8}\mathbf{=}\mathbf{9}\phantom{\rule{0ex}{0ex}}-1+\left(-8\right)=-9$

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,

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

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

$2{x}^{2}+x+8x+4=\left(2{x}^{2}+x\right)+\left(8x+4\right)$

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

$\left(2{x}^{2}+x\right)+\left(8x+4\right)=x\left(2x+1\right)+4\left(2x+1\right)$

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:

$\mathbf{\left(}\mathbit{x}\mathbf{+}\mathbf{4}\mathbf{\right)}\mathbf{\left(}\mathbf{2}\mathbit{x}\mathbf{+}\mathbf{1}\mathbf{\right)}$

Therefore,

$2{x}^{2}+9x+4=\mathbf{\left(}\mathbit{x}\mathbf{+}\mathbf{4}\mathbf{\right)}\mathbf{\left(}\mathbf{2}\mathbit{x}\mathbf{+}\mathbf{1}\mathbf{\right)}$

It is noteworthy that the factors of a quadratic expression can be arrived at using other methods such as, completing the square method, almighty formula, and the graphical method.

## Further examples of factorising expressions

A better understanding of expressions can be achieved by practicing as many problems as possible.

Solve the following.

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

Solution:

To factorise:

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

you have to expand the expression. Thus,

$14{\left(k+1\right)}^{2}+21\left(k+1\right)=14\left(k+1\right)\left(k+1\right)+21k+21\phantom{\rule{0ex}{0ex}}\left(k+1\right)\left(k+1\right)={k}^{2}+2k+1\phantom{\rule{0ex}{0ex}}14\left(k+1\right)\left(k+1\right)+21k+21=14\left({k}^{2}+2k+1\right)+21k+21\phantom{\rule{0ex}{0ex}}14\left({k}^{2}+2k+1\right)+21k+21=14{k}^{2}+28k+14+21k+21$

Bring like terms together.

$14{k}^{2}+28k+14+21k+21=14{k}^{2}+28k+21k+14+21\phantom{\rule{0ex}{0ex}}14{k}^{2}+28k+21k+14+21=14{k}^{2}+49k+35$

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

$14{k}^{2}+49k+35=7\left(2{k}^{2}+7k+5\right)$

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

$2{k}^{2}+7k+5=\left(2k+5\right)\left(k+1\right)$

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

$7\left(2{k}^{2}+7k+5\right)=7\left(2k+5\right)\left(k+1\right)\phantom{\rule{0ex}{0ex}}14{\left(k+1\right)}^{2}+21\left(k+1\right)=7\left(2k+5\right)\left(k+1\right)$

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:

$\left(7x+4y\right)+\left(3x+11y\right)=7x+3x+4y+11y\phantom{\rule{0ex}{0ex}}7x+3x+4y+11y=10x+15y$

Now you have the sum, factorise the expression.

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

## Factorising expressions - Key takeaways

• Factorising expressions is the process of simplifying expressions which gives rise to the greatest common divisor (GCD) outside the bracket and the result inside the bracket.
• In order to factorise simple expressions, you must follow some steps.
• Linear expressions are algebraic representations where both constants and variables are all in the first power.
• Quadratic expressions are factorised by applying the sum and product rule.

#### Flashcards in Factorising expressions 3

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How do you factorise expressions?

You factorise expressions by finding the greatest common divisor (GCD) in the expression and using it to divide through the expression so that you have the GCD outside the bracket and the result (macronym) inside the bracket.

How to factorise algebraic expressions?

You factorise algebraic expressions by finding the greatest common divisor (GCD) in the algebraic expression and using it to divide through the expression so that you have the GCD outside the bracket and the result (macronym) inside the bracket.

How to factorise algebraic expressions with powers?

If an algebraic expression has a higher power, confirm if it is a quadratic expression then factorise accordingly. Other than that, find the GCD or expand where necessary before the finding the GCD and factorising accordingly.

What does it mean to factor expressions?

You factor expressions by finding the GCD between terms in the expression.

What is an example of factoring an expression?

An example of factoring an expression is from the expression 6x+3xy, 3x is a factor.

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