Manipulating Rational Expressions Revision Notes | GCSE AQA Higher Further Maths

Worked example: Simplifying an Algebraic Fraction with a Quadratic Numerator

Simplify $\frac{x^2 - x - 6}{x^2 - 9}$.

1. $x^2 - x - 6 = (x-3)(x+2)$ — Find two numbers that multiply to $-6$ and add to $-1$: those are $-3$ and $+2$. [M1]
2. $x^2 - 9 = (x-3)(x+3)$ — Difference of two squares: $a^2 - b^2 = (a-b)(a+b)$ with $a = x$, $b = 3$. [M1]
3. $\frac{(x-3)(x+2)}{(x-3)(x+3)}$ — Both numerator and denominator are now fully factorised.
4. $\frac{\cancel{(x-3)}(x+2)}{\cancel{(x-3)}(x+3)} = \frac{x+2}{x+3}$ — $(x-3)$ appears in both, so it cancels. The result is fully simplified. [A1]

Answer: $\frac{x+2}{x+3}$

Worked example: Dividing Algebraic Fractions

Simplify $\frac{x^2 - 4}{x+1} \div \frac{x-2}{x^2 + 3x + 2}$.

1. $\frac{x^2 - 4}{x+1} \times \frac{x^2 + 3x + 2}{x-2}$ — Division by a fraction is equivalent to multiplication by its reciprocal. [M1]
2. $x^2 - 4 = (x-2)(x+2)$ $x^2 + 3x + 2 = (x+1)(x+2)$ — Difference of two squares for the first; find two numbers multiplying to $2$ and adding to $3$ for the second.
3. $\frac{(x-2)(x+2)}{x+1} \times \frac{(x+1)(x+2)}{x-2}$ — All four expressions are now factorised and ready for cancellation.
4. $\frac{\cancel{(x-2)}(x+2)}{\cancel{(x+1)}} \times \frac{\cancel{(x+1)}(x+2)}{\cancel{(x-2)}} = (x+2)^2$ — $(x-2)$ and $(x+1)$ each cancel across numerator and denominator. [M1]
5. $(x+2)^2$ — Expand only if the question asks for it; leaving in factorised form is acceptable unless instructed otherwise. [A1]

Answer: $(x+2)^2$

Worked example: Adding Fractions with Linear Denominators

Write $\frac{3}{x+2} + \frac{5}{x-1}$ as a single fraction in its simplest form.

1. Common denominator: $(x+2)(x-1)$ — The denominators share no common factor, so the common denominator is their product. [M1]
2. $\frac{3(x-1)}{(x+2)(x-1)} + \frac{5(x+2)}{(x+2)(x-1)}$ — Multiply the numerator and denominator of each fraction by the other's denominator.
3. $3(x-1) = 3x - 3$ $5(x+2) = 5x + 10$ — Expand carefully, paying attention to signs.
4. $\frac{3x - 3 + 5x + 10}{(x+2)(x-1)} = \frac{8x + 7}{(x+2)(x-1)}$ — Collect like terms in the numerator. [M1]
5. $8x + 7$ does not factorise to share a factor with $(x+2)(x-1)$ — The fraction is already in its simplest form. [A1]

Answer: $\frac{8x+7}{(x+2)(x-1)}$

Worked example: Subtracting Fractions with a Quadratic Denominator

Write $\frac{4}{x^2 - 1} - \frac{1}{x+1}$ as a single fraction in its simplest form.

1. $x^2 - 1 = (x-1)(x+1)$ — Difference of two squares. Recognising this avoids creating an unnecessarily large common denominator. [M1]
2. Common denominator: $(x-1)(x+1)$ — Since $(x+1)$ is already a factor of $(x-1)(x+1)$, the common denominator is $(x-1)(x+1)$ — no need to multiply both denominators together.
3. $\frac{4}{(x-1)(x+1)} - \frac{1 \times (x-1)}{(x+1)(x-1)}$ — Only the second fraction needs scaling; multiply its numerator and denominator by $(x-1)$.
4. $\frac{4 - (x-1)}{(x-1)(x+1)} = \frac{4 - x + 1}{(x-1)(x+1)} = \frac{5 - x}{(x-1)(x+1)}$ — Distribute the minus sign across $(x-1)$: $-(x-1) = -x+1$. Sign errors here are very common. [M1]
5. $5 - x$ shares no factor with $(x-1)(x+1)$ — The fraction is fully simplified. [A1]

Answer: $\frac{5-x}{(x-1)(x+1)}$

Worked example: Combined Operation: Simplify then Subtract

Write $\frac{x^2 + 5x + 6}{x^2 - 4} - \frac{1}{x-2}$ as a single fraction in its simplest form.

1. $x^2 + 5x + 6 = (x+2)(x+3)$ $x^2 - 4 = (x+2)(x-2)$ — Factorise numerator and denominator of the first fraction before doing anything else. [M1]
2. $\frac{(x+2)(x+3)}{(x+2)(x-2)} = \frac{x+3}{x-2}$ — Cancel the common factor $(x+2)$. The expression is now much simpler to work with. [M1]
3. Both fractions now have denominator $(x-2)$ — After simplification, the denominators match, so no scaling is needed.
4. $\frac{x+3}{x-2} - \frac{1}{x-2} = \frac{x+3-1}{x-2} = \frac{x+2}{x-2}$ — Combine over the shared denominator and simplify the numerator. [M1]
5. $x+2$ and $x-2$ share no common factor — The fraction is fully simplified. [A1]

Answer: $\frac{x+2}{x-2}$