Simplifying by factorising quadratic expressions

When the numerator or denominator is a quadratic expression, factorise it before looking for common factors. Use the tools from the factorisation chapter: a common factor first, then the difference of two squares \(a^2-b^2=(a-b)(a+b)\), a perfect square \(a^2\pm2ab+b^2=(a\pm b)^2\), or the multiplication frame for \(ax^2+bx+c\).

Reversed factors. \(2-x\) and \(x-2\) are not equal, but \(2-x=-(x-2)\). Taking out \(-1\) turns a reversed bracket into one that can cancel.

Why the restrictions stay: compare values

Since \(x^2-9=(x-3)(x+3)\) and \(x^2+x-6=(x+3)(x-2)\), the two forms agree for every allowed value. At \(x=-3\) the original fraction has no value even though the simplified form does, so \(x\ne-3\) is still stated.

Method

1. Take out any common number or letter factor first.
2. Factorise each quadratic completely.
3. Rewrite a reversed bracket using \(b-a=-(a-b)\).
4. Cancel the common brackets.
5. State every value that makes a factor of the original denominator zero.

Example

Simplify \(\dfrac{2x^2-8}{x^2+3x-10}\).

\(\begin{aligned}\dfrac{2x^2-8}{x^2+3x-10}&=\dfrac{2(x^2-4)}{(x+5)(x-2)} \\ &=\dfrac{2(x-2)(x+2)}{(x+5)(x-2)} \\ &=\dfrac{2(x+2)}{x+5},\qquad x\ne2,\ x\ne-5\end{aligned}\)

Simplify \(\dfrac{9-3x}{x^2-9}\).

\(\begin{aligned}\dfrac{9-3x}{x^2-9}&=\dfrac{-3(x-3)}{(x-3)(x+3)} \\ &=-\dfrac{3}{x+3},\qquad x\ne3,\ x\ne-3\end{aligned}\)

Check: never cancel the \(x^2\) terms of \(\dfrac{x^2-9}{x^2+x-6}\); they are terms, not factors. When you take out \(-1\) from a reversed bracket, the minus sign must appear in the answer.