Algebraic and Partial Fractions – A Level Maths Year 2

1 Simplifying Rational Expressions

A rational expression is one polynomial divided by another, such as $\dfrac{x^2 - 9}{x^2 + x - 6}$. It behaves exactly like a numerical fraction: you simplify by cancelling common factors, and you add by finding a common denominator.

Factorise every numerator and denominator fully.
For $\times$ and $\div$: flip the second fraction if dividing, then cancel factors that appear top and bottom.
For $+$ and $-$: use the lowest common denominator, then simplify the numerator.

Worked Example 1 — Cancelling

Simplify $\dfrac{x^2 - 9}{x^2 + x - 6}$.

①Factorise: $\dfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$

②Cancel the common factor $(x + 3)$: $\dfrac{x - 3}{x - 2}$

Worked Example 2 — Dividing

Simplify $\dfrac{2x^2 + 7x + 3}{x^2 - 9} \div \dfrac{4x + 2}{x^2 - 3x}$.

①Factorise: $\dfrac{(2x+1)(x+3)}{(x-3)(x+3)} \div \dfrac{2(2x+1)}{x(x-3)}$

②Flip and multiply: $\dfrac{(2x+1)(x+3)}{(x-3)(x+3)} \times \dfrac{x(x-3)}{2(2x+1)}$

③Cancel $(2x+1)$, $(x+3)$ and $(x-3)$: $\dfrac{x}{2}$

Worked Example 3 — Subtracting

Write $\dfrac{2}{x+1} - \dfrac{3}{x^2 - 1}$ as a single fraction in its simplest form.

①$x^2 - 1 = (x-1)(x+1)$, so the common denominator is $(x-1)(x+1)$.

②$\dfrac{2(x-1)}{(x-1)(x+1)} - \dfrac{3}{(x-1)(x+1)} = \dfrac{2x - 2 - 3}{(x-1)(x+1)}$

③$= \dfrac{2x - 5}{(x-1)(x+1)}$

Using $(x+1)(x^2-1)$ as the denominator also works, but you then have to cancel at the end. The lowest common denominator saves work.

Only cancel factors. In $\dfrac{x + 3}{x + 5}$ nothing cancels — the $x$s are terms, not factors.

2 Algebraic Division and Improper Fractions

A rational expression is improper if the degree of the numerator is greater than or equal to the degree of the denominator (like $\tfrac{7}{3}$), and proper if it is less (like $\tfrac{2}{3}$). Division turns an improper fraction into a polynomial plus a proper fraction.

The division identity

$\mathrm{F}(x) \equiv \mathrm{Q}(x) \times \mathrm{divisor} + \mathrm{R}(x)$ so $\dfrac{\mathrm{F}(x)}{\mathrm{divisor}} \equiv \mathrm{Q}(x) + \dfrac{\mathrm{R}(x)}{\mathrm{divisor}}$
where the degree of $\mathrm{R}$ is less than the degree of the divisor

Worked Example 4 — Dividing by a linear expression

Express $\dfrac{x^3 + 2x^2 - 5x + 1}{x - 2}$ in the form $ax^2 + bx + c + \dfrac{d}{x - 2}$.

①$x^3 \div x = x^2$. Subtract $x^2(x-2) = x^3 - 2x^2$ to leave $4x^2 - 5x + 1$.

②$4x^2 \div x = 4x$. Subtract $4x^2 - 8x$ to leave $3x + 1$.

③$3x \div x = 3$. Subtract $3x - 6$ to leave $7$.

④$\dfrac{x^3 + 2x^2 - 5x + 1}{x - 2} \equiv x^2 + 4x + 3 + \dfrac{7}{x - 2}$

Check with the remainder theorem: substituting $x = 2$ into the numerator gives $8 + 8 - 10 + 1 = 7$ ✓

Worked Example 5 — Dividing by a quadratic

Divide $2x^3 + 3x^2 - x + 4$ by $x^2 + 1$.

①$2x^3 \div x^2 = 2x$. Subtract $2x(x^2 + 1) = 2x^3 + 2x$ to leave $3x^2 - 3x + 4$.

②$3x^2 \div x^2 = 3$. Subtract $3x^2 + 3$ to leave $-3x + 1$.

③The remainder $-3x + 1$ has degree $1 < 2$, so stop.

④$\dfrac{2x^3 + 3x^2 - x + 4}{x^2 + 1} \equiv 2x + 3 + \dfrac{1 - 3x}{x^2 + 1}$

3 Partial Fractions with Linear Factors

Adding fractions combines them; partial fractions runs the process backwards, splitting one proper fraction into a sum of simpler ones. This is exactly what you need to integrate a rational function or to expand one as a series.

Distinct linear factors

$\dfrac{px + q}{(ax + b)(cx + d)} \equiv \dfrac{A}{ax + b} + \dfrac{B}{cx + d}$
$\dfrac{\ldots}{(ax + b)(cx + d)(ex + f)} \equiv \dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{ex + f}$

Factorise the denominator and write down the form with unknown constants.
Multiply through by the whole denominator to get an identity with no fractions.
Substitute the value of $x$ that makes each linear factor zero; each substitution kills all but one constant.
If needed, compare coefficients to find any constant that remains.

Worked Example 6 — Two linear factors

Express $\dfrac{5x + 1}{(x - 1)(x + 2)}$ in partial fractions.

①$\dfrac{5x + 1}{(x - 1)(x + 2)} \equiv \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$

②Multiply by $(x-1)(x+2)$: $5x + 1 \equiv A(x + 2) + B(x - 1)$

③$x = 1$: $6 = 3A$, so $A = 2$.

④$x = -2$: $-9 = -3B$, so $B = 3$.

⑤$\dfrac{5x + 1}{(x - 1)(x + 2)} \equiv \dfrac{2}{x - 1} + \dfrac{3}{x + 2}$

Check: at $x = 0$ the left side is $\dfrac{1}{-2} = -\tfrac12$ and the right side is $-2 + \tfrac32 = -\tfrac12$ ✓

Worked Example 7 — Three linear factors

Express $\dfrac{4x^2 - 2x - 6}{x(x - 1)(x + 3)}$ in partial fractions.

①$4x^2 - 2x - 6 \equiv A(x-1)(x+3) + Bx(x+3) + Cx(x-1)$

②$x = 0$: $-6 = A(-1)(3) = -3A$, so $A = 2$.

③$x = 1$: $4 - 2 - 6 = -4 = B(1)(4)$, so $B = -1$.

④$x = -3$: $36 + 6 - 6 = 36 = C(-3)(-4) = 12C$, so $C = 3$.

⑤$\dfrac{2}{x} - \dfrac{1}{x - 1} + \dfrac{3}{x + 3}$

Check the $x^2$ coefficients: $A + B + C = 2 - 1 + 3 = 4$ ✓

The cover-up rule. To find the constant over $(x - a)$, cover that factor in the original fraction and substitute $x = a$ into what is left. For Worked Example 6: cover $(x-1)$, put $x = 1$ into $\dfrac{5x+1}{x+2}$ to get $\dfrac{6}{3} = 2$. It is a shortcut for step ③, and only works for non-repeated linear factors.

4 Repeated Linear Factors

A squared factor needs two fractions

$\dfrac{\ldots}{(ax + b)(cx + d)^2} \equiv \dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^2}$

Do not leave out the $\dfrac{B}{cx+d}$ term. Writing only $\dfrac{A}{ax+b} + \dfrac{C}{(cx+d)^2}$ gives too few constants, and the identity cannot be satisfied in general.

Worked Example 8 — A single repeated factor

Express $\dfrac{2x + 1}{(x - 1)^2}$ in partial fractions.

①$\dfrac{2x + 1}{(x - 1)^2} \equiv \dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$, so $2x + 1 \equiv A(x - 1) + B$.

②$x = 1$: $B = 3$.

③Compare $x$ coefficients: $A = 2$.

④$\dfrac{2}{x - 1} + \dfrac{3}{(x - 1)^2}$

Worked Example 9 — Repeated and distinct together

Express $\dfrac{3x^2 - 9x - 3}{(x + 1)(x - 2)^2}$ in partial fractions.

①$3x^2 - 9x - 3 \equiv A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1)$

②$x = -1$: $3 + 9 - 3 = 9 = 9A$, so $A = 1$.

③$x = 2$: $12 - 18 - 3 = -9 = 3C$, so $C = -3$.

④No value of $x$ isolates $B$, so compare $x^2$ coefficients: $3 = A + B$, so $B = 2$.

⑤$\dfrac{1}{x + 1} + \dfrac{2}{x - 2} - \dfrac{3}{(x - 2)^2}$

Comparing the $x^2$ terms is usually quickest for the last constant. Alternatively substitute an easy value such as $x = 0$: $-3 = 4A - 2B + C = 4 - 2B - 3$, giving $B = 2$ ✓

5 Partial Fractions of an Improper Fraction

If the numerator's degree is at least the denominator's, there is a polynomial part as well. Either divide first and then split the proper remainder, or include the polynomial part in the identity from the start.

Worked Example 10 — Divide, then split

Express $\dfrac{2x^2 - x - 4}{(x - 1)(x + 2)}$ in the form $P + \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$.

①The denominator is $x^2 + x - 2$, degree $2$ — the same as the numerator, so the fraction is improper.

②$2x^2 - x - 4 = 2(x^2 + x - 2) + (-3x)$, so the fraction is $2 + \dfrac{-3x}{(x-1)(x+2)}$.

③$-3x \equiv A(x + 2) + B(x - 1)$. $x = 1$: $-3 = 3A$, $A = -1$. $x = -2$: $6 = -3B$, $B = -2$.

④$\dfrac{2x^2 - x - 4}{(x - 1)(x + 2)} \equiv 2 - \dfrac{1}{x - 1} - \dfrac{2}{x + 2}$

Quick way to find $P$: the polynomial part of a (quadratic)/(quadratic) is the ratio of the $x^2$ coefficients, here $\tfrac{2}{1} = 2$.

6 Quick Reference

Simplify

Factorise fully, then cancel factors.

Divide fractions

Flip the second, then multiply.

Improper

Degree of top $\geqslant$ degree of bottom: divide first.

Division identity

$\mathrm{F} \equiv \mathrm{Q} \times \mathrm{divisor} + \mathrm{R}$.

Linear factors

One constant over each factor.

Squared factor

$\dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}$.

Finding constants

Substitute roots; compare coefficients for the rest.

Check

Substitute $x = 0$ into both sides.

7 Practice Questions

Question 1

Simplify $\dfrac{x^2 + 5x + 6}{x^2 - 4}$.

▶ Show solution

$\dfrac{(x + 2)(x + 3)}{(x - 2)(x + 2)} = \dfrac{x + 3}{x - 2}$

Question 2

Write $\dfrac{1}{x - 3} - \dfrac{2}{x + 1}$ as a single fraction.

▶ Show solution

$\dfrac{(x + 1) - 2(x - 3)}{(x - 3)(x + 1)} = \dfrac{x + 1 - 2x + 6}{(x - 3)(x + 1)} = \dfrac{7 - x}{(x - 3)(x + 1)}$

Question 3

Find the quotient and remainder when $x^3 - 3x^2 + 4x - 5$ is divided by $(x - 1)$.

▶ Show solution

$x^3 \div x = x^2$; subtract $x^3 - x^2$ to leave $-2x^2 + 4x - 5$.

$-2x^2 \div x = -2x$; subtract $-2x^2 + 2x$ to leave $2x - 5$.

$2x \div x = 2$; subtract $2x - 2$ to leave $-3$.

Quotient $x^2 - 2x + 2$, remainder $-3$. (Check: $1 - 3 + 4 - 5 = -3$ ✓)

Question 4

Express $\dfrac{x + 7}{x^2 - x - 6}$ in partial fractions.

▶ Show solution

$x^2 - x - 6 = (x - 3)(x + 2)$, and $x + 7 \equiv A(x + 2) + B(x - 3)$.

$x = 3$: $10 = 5A$, $A = 2$. $x = -2$: $5 = -5B$, $B = -1$.

$\dfrac{2}{x - 3} - \dfrac{1}{x + 2}$

Question 5

Express $\dfrac{5}{(2x - 1)(x + 2)}$ in partial fractions.

▶ Show solution

$5 \equiv A(x + 2) + B(2x - 1)$.

$x = \tfrac12$: $5 = \tfrac52 A$, so $A = 2$. $x = -2$: $5 = -5B$, so $B = -1$.

$\dfrac{2}{2x - 1} - \dfrac{1}{x + 2}$

Question 6

Express $\dfrac{2x^2 + 6x - 2}{x(x + 1)(x - 2)}$ in partial fractions.

▶ Show solution

$2x^2 + 6x - 2 \equiv A(x + 1)(x - 2) + Bx(x - 2) + Cx(x + 1)$

$x = 0$: $-2 = -2A$, $A = 1$.

$x = -1$: $2 - 6 - 2 = -6 = B(-1)(-3) = 3B$, $B = -2$.

$x = 2$: $8 + 12 - 2 = 18 = C(2)(3) = 6C$, $C = 3$.

$\dfrac{1}{x} - \dfrac{2}{x + 1} + \dfrac{3}{x - 2}$ (Check: $1 - 2 + 3 = 2$ ✓ matches the $x^2$ coefficient.)

Question 7

Express $\dfrac{x^2 - 2x + 17}{(x + 3)(x - 1)^2}$ in partial fractions.

▶ Show solution

$x^2 - 2x + 17 \equiv A(x - 1)^2 + B(x + 3)(x - 1) + C(x + 3)$

$x = -3$: $9 + 6 + 17 = 32 = 16A$, $A = 2$.

$x = 1$: $1 - 2 + 17 = 16 = 4C$, $C = 4$.

$x^2$ coefficients: $1 = A + B$, so $B = -1$.

$\dfrac{2}{x + 3} - \dfrac{1}{x - 1} + \dfrac{4}{(x - 1)^2}$

Question 8

Express $\dfrac{3x^2 + 3}{(x - 1)(x + 2)}$ in the form $P + \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$.

▶ Show solution

Both have degree $2$, so $P = \tfrac{3}{1} = 3$.

$3x^2 + 3 \equiv 3(x - 1)(x + 2) + A(x + 2) + B(x - 1)$

$x = 1$: $6 = 3A$, $A = 2$. $x = -2$: $15 = -3B$, $B = -5$.

$3 + \dfrac{2}{x - 1} - \dfrac{5}{x + 2}$ (Check at $x = 0$: $\tfrac{3}{-2} = -1.5$ and $3 - 2 - 2.5 = -1.5$ ✓)

Question 9

(a) Express $\dfrac{2}{r(r + 2)}$ in partial fractions. (b) Hence find the exact value of $\dfrac{2}{1 \times 3} + \dfrac{2}{2 \times 4} + \dfrac{2}{3 \times 5} + \dfrac{2}{4 \times 6}$.

▶ Show solution

(a) $2 \equiv A(r + 2) + Br$. $r = 0$: $A = 1$. $r = -2$: $B = -1$. So $\dfrac{1}{r} - \dfrac{1}{r + 2}$.

(b) The sum is $\left(1 - \tfrac13\right) + \left(\tfrac12 - \tfrac14\right) + \left(\tfrac13 - \tfrac15\right) + \left(\tfrac14 - \tfrac16\right)$.

The $\tfrac13$ and $\tfrac14$ terms cancel, leaving $1 + \tfrac12 - \tfrac15 - \tfrac16 = \dfrac{30 + 15 - 6 - 5}{30} = \dfrac{34}{30} = \dfrac{17}{15}$.

Question 10

$\mathrm{f}(x) = \dfrac{x^2 + 4x + 6}{(x + 2)(x + 1)^2}$, $x > -1$.

(a) Express $\mathrm{f}(x)$ in partial fractions. (b) Prove that $\mathrm{f}(x) > 0$ for all $x > -1$. (c) Using the result $\dfrac{\mathrm{d}}{\mathrm{d}x}(x + a)^{n} = n(x + a)^{n-1}$, find $\mathrm{f}'(x)$ and hence the gradient of $y = \mathrm{f}(x)$ where it crosses the $y$-axis.

▶ Show solution

(a) $x^2 + 4x + 6 \equiv A(x + 1)^2 + B(x + 2)(x + 1) + C(x + 2)$

$x = -2$: $4 - 8 + 6 = 2 = A$. $x = -1$: $1 - 4 + 6 = 3 = C$. $x^2$: $1 = A + B$, so $B = -1$.

$\mathrm{f}(x) = \dfrac{2}{x + 2} - \dfrac{1}{x + 1} + \dfrac{3}{(x + 1)^2}$

(b) Numerator: $x^2 + 4x + 6 = (x + 2)^2 + 2 > 0$ for all $x$.

Denominator: for $x > -1$, $x + 2 > 0$ and $(x + 1)^2 > 0$, so it is positive.

Positive $\div$ positive is positive, so $\mathrm{f}(x) > 0$. $\blacksquare$

(c) $\mathrm{f}(x) = 2(x + 2)^{-1} - (x + 1)^{-1} + 3(x + 1)^{-2}$, so

$\mathrm{f}'(x) = -\dfrac{2}{(x + 2)^2} + \dfrac{1}{(x + 1)^2} - \dfrac{6}{(x + 1)^3}$

At $x = 0$: $\mathrm{f}'(0) = -\tfrac{2}{4} + 1 - 6 = -\dfrac{11}{2}$.

The partial-fraction form makes this easy — differentiating the original quotient directly would be much harder.

Algebraic and Partial Fractions · A Level Maths Year 2 · Created with MathJax