Multiplying & Dividing Algebraic Fractions (Edexcel IGCSE Maths)

Spec point — 2.2C

Multiplying & dividing algebraic fractions

How do I multiply algebraic fractions?

• STEP 1 Simplify both fractions first by fully factorising

  • E.g. $\frac{x}{3 x + 6} \times \frac{2 x + 4}{x + 7} = \frac{x}{3 ( x + 2 )} \times \frac{2 ( x + 2 )}{x + 7}$

• STEP 2 Cancel any common factors on top and bottom (from either fraction)

  • E.g. $\frac{x}{3 \cancel{( x + 2 )} } \times \frac{2 \cancel{( x + 2 )} }{x + 7} = \frac{x}{3} \times \frac{2}{x + 7}$

• STEP 3 Multiply the tops together Multiply the bottoms together

  • E.g. $\frac{2 x}{3 ( x + 7 )}$

• STEP 4 Check for any further factorising and cancelling

  • E.g. $\frac{2 x}{3 ( x + 7 )}$ has no common factors so is in its simplest form

Worked Example 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

$2 x^{2} - 3 x - 8 x + 12 x ( 2 x - 3 ) - 4 ( 2 x - 3 ) ( 2 x - 3 ) ( x - 4 )$

$x^{2}+3x-28$

• Two numbers that multiply to give -28 and add to give 3
• 7 and -4

$( x + 7 ) ( x - 4 )$

$x^{2}-5x-14$

• Two numbers that multiply to give -14 and add to give -5
• 2 and -7

$( x + 2 ) ( x - 7 )$

$9-4x^{2}$

• Difference of two squares
• $3^{2} - ( 2 x )^{2}$

$( 3 - 2 x ) ( 3 + 2 x )$

Write the expression using the factors

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

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

$\frac{- \cancel{( 3 - 2 x )} \cancel{( x - 4 )} }{( x + 7 ) \cancel{( x - 4 )} } \times \frac{( x + 2 ) ( x - 7 )}{ \cancel{( 3 - 2 x )} ( 3 + 2 x )} \frac{- 1}{( x + 7 )} \times \frac{( x + 2 ) ( x - 7 )}{( 3 + 2 x )}$

Write as a single fraction

• $( x + 7 )$ and $( x - 7 )$ are not common factors
• You can multiply $( x - 7 )$ by -1 to get rid of the negative sign in front of the front
• Leave the expressions factorised

$\frac{- ( x + 2 ) ( x - 7 )}{( x + 7 ) ( 3 + 2 x )} \text{or} \frac{( x + 2 ) ( 7 - x )}{( x + 7 ) ( 3 + 2 x )}$

How do I divide algebraic fractions?

• Flip (find the reciprocal of) the second fraction and replace ÷ with ×

  • So $\div\frac{a}{b}$ becomes $\times\frac{b}{a}$
  • E.g. $\frac{3x-12}{x}\div\frac{2x+8}{x+3}=\frac{3x-12}{x}\times\frac{x+3}{2x+8}$

• Then follow the same rules for multiplying two fractions

Worked Example 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)

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

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

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

Multiply the remaining numerators and denominators together

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

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

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