2.8 Rational expressions and partial fractions — Practise

1. Simplified

Answer: Factorise the top and the bottom: $x^2 - 1 = (x - 1)(x + 1)$ and $x^2 + 3x + 2 = (x + 1)(x + 2)$. Cancel the common factor $x + 1$: the expression is $\dfrac{x - 1}{x + 2}$, valid for $x \neq -2$. The cancelled factor matters: $x = -1$ made the original undefined.

2. Simplified

Answer: Factorise: $x^2 - 9 = (x - 3)(x + 3)$ and $x^2 - x - 6 = (x - 3)(x + 2)$. Cancel $x - 3$ to leave $\dfrac{x + 3}{x + 2}$. The original expression is undefined where $x = 3$, even though the simplified form is defined there.

3. Simplified

Answer: Factorise: $x^2 + 5x + 6 = (x + 2)(x + 3)$ and $x^2 - 4 = (x - 2)(x + 2)$. Cancel $x + 2$, giving $\dfrac{x + 3}{x - 2}$, valid for $x \neq 2$.

4. Simplified, with the excluded value

Answer: Factorise the numerator as a difference of two squares: $x^2 - 4 = (x - 2)(x + 2)$. Cancel $x - 2$ to leave $x + 2$. The original denominator is zero when $x = 2$, so the expression is defined only for $x \neq 2$.

5. A single fraction

Answer: The common denominator is $x(x + 1)$. Put each fraction over it: $\dfrac{2(x + 1) + 3x}{x(x + 1)} = \dfrac{5x + 2}{x(x + 1)}$. The expression is defined for $x \neq 0$ and $x \neq -1$.

6. A single fraction

Answer: The common denominator is $(x - 1)(x + 1)$. Then $\dfrac{(x + 1) - (x - 1)}{(x - 1)(x + 1)} = \dfrac{2}{x^2 - 1}$. The denominator is a difference of two squares.

7. Partial fractions

Answer: Write $\dfrac{3x + 2}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$. Then $3x + 2 = A(x + 2) + B(x - 1)$. Substituting $x = 1$ gives $5 = 3A$, so $A = \frac{5}{3}$. Substituting $x = -2$ gives $-4 = -3B$, so $B = \frac{4}{3}$. So the fractions are $\dfrac{5}{3(x - 1)} + \dfrac{4}{3(x + 2)}$.

8. Partial fractions

Answer: A repeated factor needs one fraction for each power: $\dfrac{2x + 3}{(x - 1)^2} = \dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$. Then $2x + 3 = A(x - 1) + B$. Comparing constants gives $B = 5$; comparing $x$ terms gives $A = 2$. So the fractions are $\dfrac{2}{x - 1} + \dfrac{5}{(x - 1)^2}$.

9. Partial fractions

Answer: Write $\dfrac{6x^2 + 5x - 5}{(x - 1)(x + 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2}$. Substitute the three roots of the denominator. At $x = 1$: $6 + 5 - 5 = 6$ and $(1 + 1)(1 + 2) = 6$, so $A = 1$. At $x = -1$: $6 - 5 - 5 = -4$ and $(-1 - 1)(-1 + 2) = -2$, so $B = 2$. At $x = -2$: $24 - 10 - 5 = 9$ and $(-2 - 1)(-2 + 1) = 3$, so $C = 3$. So the fractions are $\dfrac{1}{x - 1} + \dfrac{2}{x + 1} + \dfrac{3}{x + 2}$.

10. Partial fractions

Answer: Write the fractions with constant numerators: $\dfrac{7x + 4}{(2x + 1)(x + 2)} = \dfrac{A}{2x + 1} + \dfrac{B}{x + 2}$. Then $7x + 4 = A(x + 2) + B(2x + 1)$. At $x = -2$: $-10 = -3B$, so $B = \frac{10}{3}$. At $x = -\frac{1}{2}$: $-\frac{7}{2} + 4 = \frac{1}{2} = A \times \frac{3}{2}$, so $A = \frac{1}{3}$. So the fractions are $\dfrac{1}{3(2x + 1)} + \dfrac{10}{3(x + 2)}$.

11. The excluded values

Answer: The expression is not defined when the denominator is zero. $(x - 2)(x + 3) = 0$ when $x = 2$ or $x = -3$. So $x = 2$ and $x = -3$ are excluded.

12. Partial fractions

Answer: Write $\dfrac{3x^2 + 2x + 1}{(x - 1)(x + 2)^2} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} + \dfrac{C}{(x + 2)^2}$. Then $3x^2 + 2x + 1 = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1)$. At $x = 1$: $3 + 2 + 1 = 6 = 9A$, so $A = \frac{2}{3}$. At $x = -2$: $12 - 4 + 1 = 9 = -3C$, so $C = -3$. Comparing $x^2$ coefficients: $3 = A + B$, so $B = 3 - \frac{2}{3} = \frac{7}{3}$. So the fractions are $\dfrac{2}{3(x - 1)} + \dfrac{7}{3(x + 2)} - \dfrac{3}{(x + 2)^2}$.