Adding and subtracting with quadratic denominators

Factorise every quadratic denominator first. The LCD contains each different factor, to the highest power in which it appears (see the next guide for repeated factors). If a fraction can be simplified on its own, simplify it before combining.

Why: \(x^2-36=(x-6)(x+6)\) already contains the factor \(x+6\), so \((x-6)(x+6)\) is a common denominator for \(\dfrac{3}{x^2-36}\) and \(\dfrac{1}{x+6}\). Multiplying the two full denominators, \((x^2-36)(x+6)\), also works but gives a longer calculation and an answer that must be simplified later.

Method

1. Factorise each denominator, and each numerator if it may cancel.
2. Simplify any single fraction that has a common factor.
3. Build the LCD from the different factors.
4. Rewrite, combine, expand and collect as before.
5. Factorise the final numerator to check for cancelling, and state the restrictions.

Example

Express \(\dfrac{3}{x^2-36}+\dfrac{1}{x+6}\) as a single fraction.

\(\begin{aligned}\dfrac{3}{x^2-36}+\dfrac{1}{x+6}&=\dfrac{3}{(x-6)(x+6)}+\dfrac{x-6}{(x-6)(x+6)} \\ &=\dfrac{3+x-6}{(x-6)(x+6)} \\ &=\dfrac{x-3}{(x-6)(x+6)},\qquad x\ne6,\ x\ne-6\end{aligned}\)

Simplifying first: express \(\dfrac{3x-6}{x^2+x-6}-\dfrac{1}{x+3}\) as a single fraction.

\(\begin{aligned}\dfrac{3x-6}{x^2+x-6}-\dfrac{1}{x+3}&=\dfrac{3(x-2)}{(x+3)(x-2)}-\dfrac{1}{x+3} \\ &=\dfrac{3}{x+3}-\dfrac{1}{x+3} \\ &=\dfrac{2}{x+3},\qquad x\ne2,\ x\ne-3\end{aligned}\)

Check: after combining, factorise the numerator and look for a factor to cancel. The restriction \(x\ne2\) stays even though \(x-2\) cancelled.