Simplifying by recognising special products

The most effective way to simplify an algebraic fraction is to recognise the special products “hidden” in its polynomials. Once the numerator and denominator have been factorised, the common factors cancel and the expression becomes much simpler.

Example

$\frac{8x^3+12x^2+6x+1}{4x^2+4x+1} - \frac{4x^2+4x+1}{8x^3+12x^2+6x+1} = 0$ We recognise the cubes and squares of a binomial: $8x^3+12x^2+6x+1 = (2x+1)^3, \qquad 4x^2+4x+1 = (2x+1)^2.$ E.C.: $x\neq -\dfrac{1}{2}$. $\begin{aligned} \frac{(2x+1)^3}{(2x+1)^2} - \frac{(2x+1)^2}{(2x+1)^3} &= 0 \\[4pt] (2x+1) - \frac{1}{2x+1} &= 0 \end{aligned}$

After simplification the initial expression, seemingly complicated, reduces to an elementary relation between $2x+1$ and its reciprocal.

Links

Topics: Algebraic fractions
Concepts: Existence conditions · Algebraic fraction · Special products · Simplification
Skills: Factorising · Simplifying