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Factorise the following expressions.

Note: Use an asterisk (*) for multiplication and a hat (^) for raising to a power when entering your answers. For example, $xy$ should be written as $x$*$y$, and $a(b+c)$ should be typed as $a$*$(b+c)$ and $x^2$ should be typed as $x$^2.

Also, the steps provided under 'Show steps' give worked solutions for you to study. Once the concepts are clear, it may be useful to click 'Try another question like this one' to refresh the numbers and complete the question without checking the steps.

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Follow the steps outlined above to help you factorise the questions.

Click on 'Try another question like this' to try other questions using different numbers.

Alternatively, click on this link to watch a video on factorising expressions.

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Factorising polynomials using the highest common factor.

Adapted from 'Factorisation' by Steve Kilgallon.

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$\\var{c[0]}-\\var{c[1]}x^2=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $2$. The HCF, therefore, is placed outside of the brackets: $2(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $2$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$2\\big(\\frac{\\var{c[0]}}{2}-\\frac{\\var{c[1]}}{2}x^2\\big)=2\\big(\\simplify{{c[0]}/2}-\\simplify{{c[1]}/2}x^2\\big)$

Your answer must be factorised

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$\\var{c[2]}ab+\\var{c[3]}bc=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $2b$. The HCF, therefore, is placed outside of the brackets: $2b(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by two $2b$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$2b\\big(\\frac{\\var{c[2]}ab}{2b}+\\frac{\\var{c[3]}bc}{2b}\\big)=2b\\big(\\simplify{({c[2]}/2)*a}+\\simplify{({c[3]}/2)*c}\\big)$

Your answer must be factorised

$\\var{c[4]}a^2+\\var{c[5]}ab=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $2a$. The HCF, therefore, is placed outside of the brackets: $2a(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $2a$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$2a\\big(\\frac{\\var{c[4]}a^2}{2a}+\\frac{\\var{c[5]}ab}{2a}\\big)=2a\\big(\\simplify{({c[4]}/2)*a}+\\simplify{({c[5]}/2)*b}\\big)$

Your answer must be factorised

$pq^3-p^3q=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $pq$. The HCF, therefore, is placed outside of the brackets: $pq(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $pq$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$pq\\big(\\frac{pq^3}{pq}-\\frac{p^3q}{pq}\\big)=pq(q^2-p^2)$

Your answer must be factorised

$\\var{c2[0]}x^2y+\\var{c2[1]}xy^4=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $3xy$. The HCF, therefore, is placed outside of the brackets: $3xy(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $3xy$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$3xy\\big(\\frac{\\var{c2[0]}x^2y}{3xy}+\\frac{\\var{c2[1]}xy^4}{3xy}\\big)=3xy\\big(\\simplify{({c2[0]}/3)*x}+\\simplify{({{c2}[1]}/3)*y^3}\\big)$

Your answer must be factorised

$\\var{c3[0]}p^3q-\\var{c3[1]}p^2q^2+\\var{c3[2]}pq^3=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $2pq$. The HCF, therefore, is placed outside of the brackets: $2pq(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $2pq$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$2pq\\big(\\frac{\\var{c3[0]}p^3q}{2pq}-\\frac{\\var{c3[1]}p^2q^2}{2pq}+\\frac{\\var{c3[2]}pq^3}{2pq}\\big)=2pq\\big(\\simplify{({c3[0]}/2)p^2}-\\simplify{({c3[1]}/2)*p q}+\\simplify{({c3[2]}/2)*q^2}\\big)$

Your answer must be factorised

$\\var{c2[2]}lm^2-\\var{c2[3]}l^3m^3+\\var{c2[4]}l^2m^4=$

", "steps": [{"prompt": "

First, find the highest common factor between the two coefficients. In this case it is $3lm^2$. The HCF, therefore, is placed outside of the brackets: $3lm^2(\\;\\;\\;\\;)$

Next, put into the brackets the terms which, when multiplied by $2pq$, result in the the original expression.

To find this, work backwards by dividing each term by the common factor.

$3lm^2\\big(\\frac{\\var{c2[2]}lm^2}{3lm^2}-\\frac{\\var{c2[3]}l^3m^3}{3lm^2}+\\frac{\\var{c2[4]}l^2m^4}{3lm^2}\\big)=3lm^2\\big(\\simplify{({c2[2]}/3)}-\\simplify{({c2[3]}/3)*l^2m}+\\simplify{({c2[4]}/3)*lm^2}\\big)$

Your answer must be factorised

$\\var{firstsquare}-\\var{secondsquare}x^2$ =

", "steps": [{"prompt": "

An expression such this is known as 'the difference of two squares'. This occurs when one square number or term in subtracted from another. The general pattern observed is the following:

$a^2-b^2=(a+b)(a-b)$

This because, when the brackets on the right hand side are expanded, the terms simplify:

$(a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2$

There may be other varables involved, but the same still applies to the term as a whole. For example:

$a^2x^4-b^2y^6=(ax^2+by^3)(ax^2-by^3)$

Assigning values to coefficients, another example might be:

$9x^2-16y^4=(3x+4y^2)(3x-4y^2)$

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$\\simplify{{qa}*x^2+{qb}*x+{qc}}$ =

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When doing the opposite of factorisation, i.e. expanding brackets, it is sometimes possible to reveal a quadratic expression such as the one in this question from the product of two binomials where the highest power of the variable is $1$. It follows, therefore, that some quadratic expressions can be factorised into the product of two binomials. When this is possible, we say that the quadratic expression can be 'factorised with integers'.

For example, $-6x^2+x+2=(2x+1)(-3x+2)$

In more general terms, the quadratic expression can be shown as:

$ax^2+bx+c$ and the product of binomials could be $(dx+e)(fx+g)$

When expanding the brackets of the binomial products, the quadratic is formed as:

$(d\\times f)x^2+((d\\times g)+(e\\times f))x+(e\\times g)$

Factorising the quadratic with integers, therefore, requires the insight of reverse engineering the products of the coefficients to reflect the pattern shown above.

In this question, the coefficients of the quadratic expression can be analysed as follows:

$a=(d\\times f)=\\var{qa},\\;\\;\\;b=((d\\times g)+(e\\times f))=\\var{qb},\\;\\;\\;c=(e\\times g)=\\var{qc}$

Look for all possible factors of these coefficients and try different variations until you find those which satisfy all conditions.

The result should be:

$d=\\var{q1},\\;\\;\\;e=\\var{q2}\\;\\;\\;f=\\var{q3}\\;\\;\\;g=\\var{q4}$

So that,

$(\\var{q1}\\times\\var{q3})x^2+((\\var{q1}\\times\\var{q4})+(\\var{q2}\\times\\var{q3}))x+(\\var{q2}\\times\\var{q4})$

$\\therefore,\\;\\;\\;\\simplify{{qa}*x^2+{qb}*x+{qc}}=(\\simplify{{q1}x+{q2}})(\\simplify{{q3}*x+{q4}})$

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Coefficients in e,f (HCF: 3)

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Part f (HCF:2)

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Coefficients in a,b,c (HCF: 2)

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