Multiplying & Dividing Algebraic Fractions (WJEC Eduqas GCSE Maths: Higher): Revision Note

Write the following as a single fraction in its simplest form

$\frac{2x^2-11x+12}{x^2+3x-28}\times \frac{x^2-5x-14}{9-4x^2}$

Answer:

Factorise the four quadratics

$2x^2-11x+12$

• Two numbers that multiply to give 2×12=24 and add to give -11

• -3 and -8

$\begin{array}{l}2x^2-3x-8x+12\\ x\left(2x-3\right)-4\left(2x-3\right)\\ \left(2x-3\right)\left(x-4\right)\end{array}$

$x^2+3x-28$

• Two numbers that multiply to give -28 and add to give 3

• 7 and -4

$\left(x+7\right)\left(x-4\right)$

$x^2-5x-14$

• Two numbers that multiply to give -14 and add to give -5

• 2 and -7

$\left(x+2\right)\left(x-7\right)$

$9-4x^2$

• Difference of two squares

• $3^2-{\left(2x\right)}^2$

$\left(3-2x\right)\left(3+2x\right)$

Write the expression using the factors

$\frac{\left(2x-3\right)\left(x-4\right)}{\left(x+7\right)\left(x-4\right)}\times \frac{\left(x+2\right)\left(x-7\right)}{\left(3-2x\right)\left(3+2x\right)}$

Rewrite $2x-3$ as $-\left(3-2x\right)$ and cancel common factors

$\begin{array}{l}\frac{-\cancel{\left(3-2x\right)}\cancel{\left(x-4\right)}}{\left(x+7\right)\cancel{\left(x-4\right)}}\times \frac{\left(x+2\right)\left(x-7\right)}{\cancel{\left(3-2x\right)}\left(3+2x\right)}\\ \frac{-1}{\left(x+7\right)}\times \frac{\left(x+2\right)\left(x-7\right)}{\left(3+2x\right)}\end{array}$

Write as a single fraction

• $\left(x+7\right)$ and $\left(x-7\right)$ are not common factors

• You can multiply $\left(x-7\right)$ by -1 to get rid of the negative sign in front of the front

• Leave the expressions factorised

$\frac{-\left(x+2\right)\left(x-7\right)}{\left(x+7\right)\left(3+2x\right)}\mathrm{or}\frac{\left(x+2\right)\left(7-x\right)}{\left(x+7\right)\left(3+2x\right)}$

Divide $\frac{x+3}{x-4}$ by $\frac{2x+6}{x^2-16}$, giving your answer as a simplified fraction.

Answer:

Division is the same as multiplying by the reciprocal (the fraction flipped)

$\frac{x+3}{x-4}÷\frac{2x+6}{x^2-16}=\frac{x+3}{x-4}\times \frac{x^2-16}{2x+6}$

Factorise all numerators and denominators to see which factors cancel out
You need to use the difference of two squares, $x^2-4^2=\left(x-4\right)\left(x+4\right)$

$\frac{x+3}{x-4}\times \frac{x^2-16}{2x+6}=\frac{\cancel{x+3}}{\cancel{x-4}}\times \frac{\cancel{\left(x-4\right)}\left(x+4\right)}{2\cancel{\left(x+3\right)}}$

Multiply the remaining numerators and denominators together

$\frac{1\times \left(x+4\right)}{1\times 2}=\frac{x+4}{2}$

Check to see if you missed any terms that are the same on the top and bottom that could be cancelled

$\frac{x+4}{2}$ is already in its simplest form

$\frac{\mathit{x}\mathbf{+}\mathbf{4}}{\mathbf{2}}$