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A question to practice simplifying fractions with the use of factorisation (for binomial and quadratic expressions).

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Simplify the following algebraic expressions.

Note: Although the question may accept coefficients in their decimal forms, it would be more appropriate to keep them in their most simplified fraction forms.

", "advice": "

Click 'Try another question like this one' if you need more practice.

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$\\simplify{({n1}{a1}x^2+{n1}{a2}x)/({d1}{a1}x+{d1}{a2})}$

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${\\simplify{({n1}{a1}x^2+{n1}{a2}x)/({d1}{a1}x+{d1}{a2})}}$

Factorise the numerator and denominator so that the binomials in both are the same.

${\\big(\\frac{\\var{n1}x}{\\var{d1}}\\big)\\big(\\frac{\\var{a1}x+\\var{a2}}{\\var{a1}x+\\var{a2}}\\big)}$

The binomials cancel, leaving $x$ and its coefficient:

$\\big({\\simplify{{n1}/{d1}}}\\big)x$

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Please simplify further.

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$\\simplify{(({n2}{b1}n^({p2}+1)+{n2}{b2}n^{p2})/({n2}{b3}n^({p2}+1)+{n2}{b4}n^{p2}))}$

${\\simplify{(({n2}{b1}n^({p2}+1)+{n2}{b2}n^{p2})/({n2}{b3}n^({p2}+1)+{n2}{b4}n^{p2}))}}$

As before, factorise the numerator and denominator. This time, however, you'll notice that the factors themselves are the same.

$\\big(\\frac{\\var{n2}n}{{\\var{n2}}n}\\big)\\big(\\frac{\\var{b1}n+\\var{b2}}{\\var{b3}n+\\var{b4}}\\big)$

The factors cancel, leaving:

$\\big(\\frac{\\var{b1}n+\\var{b2}}{\\var{b3}n+\\var{b4}}\\big)$

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$\\simplify{(x^2+({c1}+{c2})x +{c1}{c2})/(x+{c1})}$

${\\simplify{(x^2+({c1}+{c2})x +{c1}{c2})/(x+{c1})}}$

Here, the quadratic expression in the numerator needs to be factorised into the product of two binomials.

$\\frac{({\\simplify{x+{c1}}})({\\simplify{x+{c2}}})}{({\\simplify{x+{c1}}})}$

You will notice that one of the binomials in the numerator is the same as the denominator, which means that they can be cancelled. This leaves the expression:

${\\simplify{x+{c2}}}$

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$\\frac{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}$

$\\frac{{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}}{{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}}$

This time there is a quadratic expression which needs to be factorised into the products of binomials in both the numerator and denominator.

$\\frac{({\\simplify{n+{e1}}})({\\simplify{n+{e2}}})}{({\\simplify{n+{e1}}})({\\simplify{n+{e3}}})}$

The repeated binomials in the numerator and denominator cancel, leaving:

$\\frac{({\\simplify{n+{e2}}})}{({\\simplify{n+{e3}}})}$

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$\\simplify{({co1}x+{co1}{f1})/(x+{f2})}\\times \\simplify{({co2}x+{co2}{f2})/(x+{f1})}$

${\\simplify{({co1}x+{co1}{f1})/(x+{f2})}}\\times{\\simplify{({co2}x+{co2}{f2})/(x+{f1})}}$

For this question, start by factorising each fraction being multiplied.

$\\big(\\var{co1}\\big)\\big(\\frac{\\simplify{x+{f1}}}{\\simplify{x+{f2}}}\\big)\\times\\big(\\var{co2}\\big)\\big(\\frac{\\simplify{x+{f2}}}{\\simplify{x+{f1}}}\\big)$

Due to the commmutative nature of multiplication, the factors can be rearranged so that potential simplification becomes easier to spot.

$\\big(\\var{co1}\\times\\var{co2}\\big)\\Big(\\frac{(\\simplify{x+{f1}})(\\simplify{x+{f2}})}{(\\simplify{x+{f2}})(\\simplify{x+{f1}})}\\Big)$

The binomial expressions in the fraction all cancel, leaving the answer as the product of the factorised coefficients:

$\\var{co1}\\times\\var{co2}=\\simplify{{co1}*{co2}}$

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