Rational expressions and algebraic divisionAQA A-Level Maths: Revision notes

A rational expression is a fraction whose numerator and denominator are polynomials, such as $\frac{x^2-9}{x^2+x-12}$. To simplify, factorise the numerator and denominator completely, then cancel common factors. Example: $\frac{x^2-9}{x^2+x-12}=\frac{(x-3)(x+3)}{(x+4)(x-3)}=\frac{x+3}{x+4}$. You may only cancel factors (things multiplied), never terms (things added). In $\frac{x+3}{x+4}$ nothing can be cancelled, and $\frac{x^2-9}{x^2+x-12}$ does not become $\frac{-9}{x-12}$. The original expression is undefined when a denominator is zero, so $\frac{x^2-9}{x^2+x-12}$ is undefined at $x=3$ and $x=-4$, even though the simplified form is defined at $x=3$.

To multiply rational expressions, factorise everything, multiply across and cancel common factors (cancelling before multiplying is easiest). To divide, multiply by the reciprocal of the second fraction. Example: $E(x)=\frac{x^2+2x-15}{x^2-4x+3}\div\frac{x+5}{x^2-1}$. $=\frac{(x+5)(x-3)}{(x-1)(x-3)}\times\frac{(x-1)(x+1)}{x+5}=x+1$. Be careful about where the original is undefined: $E$ involves division by $\frac{x+5}{x^2-1}$, which is zero at $x=-5$, and has zero denominators at $x=1$, $x=-1$ and $x=3$. So $E(x)=x+1$ except at those four values.

To add or subtract, write over a common denominator, which is usually the product of the denominators, or the lowest common multiple if they share a factor. Example: $\frac{2}{x+1}+\frac{3}{x-2}=\frac{2(x-2)+3(x+1)}{(x+1)(x-2)}=\frac{5x-1}{(x+1)(x-2)}$. Example with a factor in common: $\frac{x^2+3x-4}{x^2-1}-\frac{2}{x+1}$. Simplify first: $\frac{(x+4)(x-1)}{(x-1)(x+1)}=\frac{x+4}{x+1}$. Then $\frac{x+4}{x+1}-\frac{2}{x+1}=\frac{x+2}{x+1}$. Simplifying first keeps the algebra small. Take care to subtract the whole numerator: $\frac{a}{c}-\frac{b+d}{c}=\frac{a-b-d}{c}$.

A rational expression is improper when the degree of the numerator is at least the degree of the denominator. Divide the numerator by the linear denominator to write it as a quotient plus a remainder fraction. Example: $\frac{2x^3-5x^2+x+7}{x-2}$. Dividing, $2x^3\div x=2x^2$, leaving $-x^2+x+7$; then $-x$, leaving $-x+7$; then $-1$, leaving $5$. So $2x^3-5x^2+x+7=(x-2)(2x^2-x-1)+5$, and $\frac{2x^3-5x^2+x+7}{x-2}=2x^2-x-1+\frac{5}{x-2}.$ If the remainder is zero, the division is exact and the fraction simplifies to a polynomial, e.g. $\frac{2x^2+3x-5}{x-1}=2x+5$. Check any result by multiplying the quotient by the divisor and adding the remainder.

To solve an equation containing algebraic fractions, simplify, then multiply both sides by the denominator and solve. Examples: $\frac{x+3}{x+4}=2\Rightarrow x+3=2x+8\Rightarrow x=-5$. $\frac{x+2}{x+1}=3\Rightarrow x+2=3x+3\Rightarrow x=-\frac12$. If the equation becomes a quadratic, solve it by factorising or the formula: $\frac{2}{x+1}+\frac{3}{x-2}=1\Rightarrow5x-1=(x+1)(x-2)\Rightarrow x^2-6x-1=0$, so $x=3\pm\sqrt{10}$. Finally, check that no solution makes an original denominator zero. For an expression equal to $0$, the numerator must be zero (and the denominator not zero).