Bradley's Maths

Solution

This question is a very early-exam non-calculator question, so GCSE Paper 1 around question 1 to 4 and IGCSE Paper 2 in about the same place. Whilst simplifying algebraic fractions is a relatively low level concept, it nevertheless is important as it is a necessary pre-requisite for any other algebraic fraction work which may come up in your examination.

The most important rule: If any part of your fraction (numerator, denominator or both) factorises you MUST factorise first, before attempting any cancelling.

a) There is no factorisation involved in this one so we can go directly into cancelling. Whilst you are still learning the process I strongly recommending cancelling any algebraic elements first and the numeric elements at the end.

\begin{align*} \dfrac{25a^2b}{30a} &= \dfrac{25\cancelto{a}{a^2}b}{30\cancelto{1}{a}}\\ &= \dfrac{25ab}{30} \\ &= \dfrac{5ab}{6} \end{align*}

Final Answer:

$$\dfrac{5ab}{6}$$

b) The numerator has a factor of $p^2$ and the denominator has a factor of $2$. We must deal with these first.

\begin{align*} \dfrac{p^3+p^2}{2p+2} &=\dfrac{p^2(p+1)}{2(p+1)}\\ &=\dfrac{p^2\cancel{(p+1)}}{2\cancel{(p+1)}}\\ &=\dfrac{p^2}{2} \end{align*}

Final Answer:

$$\dfrac{p^2}{2}$$

c) In this case we need to factorise the quadratic numerator. The $x+5$ denominator is an examiner's clue as to what the factorisation may be!

\begin{align*} \dfrac{x^2-2x-35}{x+5} &=\dfrac{(x+5)(x-7)}{x+5}\\ &=\dfrac{\cancel{(x+5)}(x-7)}{\cancel{x+5}}\\ &=x-7 \end{align*}

Final Answer:

$$x-7$$

d) Both the numerator and denominator need to be factorised. Watch out for the disguised difference of two squares in the numerator.

\begin{align*} \dfrac{3x^2-27}{x^2+6x+9} &=\dfrac{3(x^2-9)}{(x+3)^2}\\ &=\dfrac{3(x-3)(x+3)}{(x+3)(x+3)}\\ &=\dfrac{3(x-3)\cancel{(x+3)}}{\cancel{(x+3)}(x+3)}\\ &=\dfrac{3(x-3)}{x+3} \end{align*}

Final Answer:

$$\dfrac{3(x-3)}{x+3}$$