2.8 Rational expressions and partial fractions — Practise with help

1. $x^2 - 5x + 6 = (x - 2)(x - 3)$.
2. $x^2 - 16 = (x - 4)(x + 4)$.
3. $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3 + 2}{6} = \dfrac{5}{6}$.

If question 3 was hard, revisit the common denominator: it is the smallest number that both denominators divide.

Factorise the top as a difference of two squares, and factorise the bottom as a quadratic with two whole-number roots.

$x^2 - 4 = (x - 2)(x + 2)$ and $x^2 + 5x + 6 = (x + 2)(x + 3)$. Cancel the common factor.

Factorise:

Cancel $x + 2$:

The original expression is excluded at $x = -2$ and $x = -3$.

The common denominator is the product $(x + 2)(x - 1)$.

Put both fractions over the common denominator: $\dfrac{3(x - 1) + 4(x + 2)}{(x + 2)(x - 1)}$.

Put both fractions over a common denominator:

The expression is excluded at $x = -2$ and $x = 1$.

Write $\dfrac{A}{x - 2} + \dfrac{B}{x + 1}$, then multiply both sides by $(x - 2)(x + 1)$.

$5x + 1 = A(x + 1) + B(x - 2)$. Substitute $x = 2$, then $x = -1$.

Write

Multiply through:

At $x = 2$: $11 = 3A$, so $A = \frac{11}{3}$. At $x = -1$: $-4 = -3B$, so $B = \frac{4}{3}$. Therefore

A squared denominator needs two terms: one with $x + 2$ and one with $(x + 2)^2$.

Write $\dfrac{A}{x + 2} + \dfrac{B}{(x + 2)^2}$, so that $4x - 1 = A(x + 2) + B$.

Write

Then

The $x$ terms give $A = 4$. Both sides must agree at every $x$; at $x = -2$, $-9 = B$. So

Check the degree first. The numerator has degree $2$, the same as the denominator, so divide before you split.

Divide: $2x^2 - x + 3 = 2(x - 1)(x + 2) + \ldots$. Work out the remainder.

First divide, since the fraction is improper. Expand $(x - 1)(x + 2) = x^2 + x - 2$, and compare leading terms: $2(x^2 + x - 2) = 2x^2 + 2x - 4$. Subtract from the numerator:

So

Now split the proper part: $-3x + 7 = A(x + 2) + B(x - 1)$. At $x = 1$: $4 = 3A$, so $A = \frac{4}{3}$. At $x = -2$: $13 = -3B$, so $B = -\frac{13}{3}$. Therefore

The common denominator is $(x - a)(x + a)$, which is $x^2 - a^2$.

The numerator is $(x + a) - (x - a) = 2a$.

Put over the common denominator:

The expression is excluded where $x^2 - a^2 = 0$, so $x = a$ and $x = -a$ are not allowed.