Mathematics 9709 · Pure Mathematics 3

Partial fractions

Partial fractions is just adding fractions in reverse. You're given the combined answer and have to split it back into simple pieces.
The whole skill is one choice: look at the bottom and pick which of four patterns to write down.
After that, finding the constants is the easy part.

The bottom picks the form. Distinct linear, repeated linear, irreducible quadratic, or improper: each has its own fixed template. Get the template right and the rest is just solving for the letters.

The four patterns (read the bottom)

Distinct linear $(ax+b)(cx+d)$ : $\dfrac{A}{ax+b}+\dfrac{B}{cx+d}$
Repeated linear $(ax+b)^2$ : $\dfrac{A}{ax+b}+\dfrac{B}{(ax+b)^2}$ (one term per power)
Irreducible quadratic $(cx^2+d)$ : $\dfrac{A}{ax+b}+\dfrac{Bx+C}{cx^2+d}$ (top must be linear)
Improper (top degree ≥ bottom): divide first (§7.1), then split the proper part.

To find the constants, multiply both sides by the whole bottom. Then plug in the root of each factor. That kills every term but one, so you read off a constant in one line. Fill any gaps left over by matching coefficients.

Why plugging in the root is the fast route

After you multiply up you get an identity like $2x-13=A(x-3)+B(2x+1)$ that is true for every $x$ . Pick $x=3$ and the $A(x-3)$ term goes to zero. That leaves a one-line equation for $B$ . Pick $x=-\tfrac12$ and the $B$ term goes to zero, giving $A$ . Each root makes one term vanish, which is much quicker than solving two equations together.

For a repeated or quadratic factor a few roots are not enough. So you finish by matching coefficients (compare the $x^2$ terms, say) to find the last letter.

Worked examples (one per case)

Worked example

Distinct linear: $\dfrac{2x-13}{(2x+1)(x-3)}$

Two distinct linear factors, so write $\dfrac{2x-13}{(2x+1)(x-3)}=\dfrac{A}{2x+1}+\dfrac{B}{x-3}$ and multiply up: $2x-13=A(x-3)+B(2x+1)$ .
Substitute the roots. Put $x=3$ : $-7=7B\Rightarrow B=-1$ . Put $x=-\tfrac12$ : $-14=-\tfrac72A\Rightarrow A=4$ .

Answer

\dfrac{4}{2x+1}-\dfrac{1}{x-3}

Worked example

Repeated linear: $\dfrac{2x-11}{(x-4)^2}$

Repeated factor needs a term for each power: $\dfrac{2x-11}{(x-4)^2}=\dfrac{A}{x-4}+\dfrac{B}{(x-4)^2}$ , so $2x-11=A(x-4)+B$ .
Put $x=4$ : $B=-3$ . There's no second root, so match the $x$ -coefficient: $A=2$ .

Answer

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

Worked example

Quadratic factor: $\dfrac{8+2x-x^2}{(x-1)(x^2+2)}$

An irreducible $x^2+2$ takes a linear top: $\dfrac{8+2x-x^2}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$ , giving $8+2x-x^2=A(x^2+2)+(Bx+C)(x-1)$ .
Put $x=1$ : $9=3A\Rightarrow A=3$ . Put $x=0$ : $8=2A-C\Rightarrow C=-2$ . Match $x^2$ : $-1=A+B\Rightarrow B=-4$ .

Answer

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

Worked example

Improper: $\dfrac{2x^2+5x-11}{x^2+2x-3}$

Same degree top and bottom — improper, so divide first (§7.1): $\dfrac{2x^2+5x-11}{x^2+2x-3}=2+\dfrac{x-5}{x^2+2x-3}$ .
Now split the proper remainder. Factorise $x^2+2x-3=(x+3)(x-1)$ and write $x-5=A(x-1)+B(x+3)$ . Put $x=1$ : $B=-1$ ; put $x=-3$ : $A=2$ .

Answer

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

Where the marks go

Common mistake

An irreducible quadratic needs a linear top

Bx+C

, not just a number. And a repeated factor needs both

\dfrac{A}{ax+b}

and

\dfrac{B}{(ax+b)^2}

. Get the template wrong and you lose the mark before you have done any arithmetic.

Common mistake

If the fraction is improper, divide first (§7.1). Set up partial fractions on an improper fraction and it never balances. The polynomial part has no place to go.

Common mistake

When you finish by matching coefficients, compare the highest power first. It usually has the fewest terms. Then work down. Start with the wrong power and you tie yourself up in needless equations.

Now you try

Split $\dfrac{2x^2+7x+8}{(1+x)(2+x)^2}$ into partial fractions. (9709/31 Oct/Nov 2022 Q10)

Your turn— tap to reveal the worked answer (9709/31 Oct/Nov 2022 Q10(a))

There's a repeated factor, so use the template $\dfrac{A}{1+x}+\dfrac{B}{2+x}+\dfrac{C}{(2+x)^2}$ . Multiply up: $2x^2+7x+8=A(2+x)^2+B(1+x)(2+x)+C(1+x)$ .

Put $x=-1$ : $A=3$ ; put $x=-2$ : $C=-2$ ; match $x^2$ : $A+B=2\Rightarrow B=-1$ .

Answer

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

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Partial fractions

Partial fractions is just adding fractions in reverse. You're given the combined answer and have to split it back into simple pieces.
The whole skill is one choice: look at the bottom and pick which of four patterns to write down.
After that, finding the constants is the easy part.

The four patterns (read the bottom)

Distinct linear $(ax+b)(cx+d)$ : $\dfrac{A}{ax+b}+\dfrac{B}{cx+d}$
Repeated linear $(ax+b)^2$ : $\dfrac{A}{ax+b}+\dfrac{B}{(ax+b)^2}$ (one term per power)
Irreducible quadratic $(cx^2+d)$ : $\dfrac{A}{ax+b}+\dfrac{Bx+C}{cx^2+d}$ (top must be linear)
Improper (top degree ≥ bottom): divide first (§7.1), then split the proper part.

Why plugging in the root is the fast route

For a repeated or quadratic factor a few roots are not enough. So you finish by matching coefficients (compare the $x^2$ terms, say) to find the last letter.

Worked examples (one per case)

Worked example

Distinct linear: $\dfrac{2x-13}{(2x+1)(x-3)}$

Two distinct linear factors, so write $\dfrac{2x-13}{(2x+1)(x-3)}=\dfrac{A}{2x+1}+\dfrac{B}{x-3}$ and multiply up: $2x-13=A(x-3)+B(2x+1)$ .
Substitute the roots. Put $x=3$ : $-7=7B\Rightarrow B=-1$ . Put $x=-\tfrac12$ : $-14=-\tfrac72A\Rightarrow A=4$ .

Answer

\dfrac{4}{2x+1}-\dfrac{1}{x-3}

Worked example

Repeated linear: $\dfrac{2x-11}{(x-4)^2}$

Repeated factor needs a term for each power: $\dfrac{2x-11}{(x-4)^2}=\dfrac{A}{x-4}+\dfrac{B}{(x-4)^2}$ , so $2x-11=A(x-4)+B$ .
Put $x=4$ : $B=-3$ . There's no second root, so match the $x$ -coefficient: $A=2$ .

Answer

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

Worked example

Quadratic factor: $\dfrac{8+2x-x^2}{(x-1)(x^2+2)}$

An irreducible $x^2+2$ takes a linear top: $\dfrac{8+2x-x^2}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$ , giving $8+2x-x^2=A(x^2+2)+(Bx+C)(x-1)$ .
Put $x=1$ : $9=3A\Rightarrow A=3$ . Put $x=0$ : $8=2A-C\Rightarrow C=-2$ . Match $x^2$ : $-1=A+B\Rightarrow B=-4$ .

Answer

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

Worked example

Improper: $\dfrac{2x^2+5x-11}{x^2+2x-3}$

Same degree top and bottom — improper, so divide first (§7.1): $\dfrac{2x^2+5x-11}{x^2+2x-3}=2+\dfrac{x-5}{x^2+2x-3}$ .
Now split the proper remainder. Factorise $x^2+2x-3=(x+3)(x-1)$ and write $x-5=A(x-1)+B(x+3)$ . Put $x=1$ : $B=-1$ ; put $x=-3$ : $A=2$ .

Answer

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

Where the marks go

Common mistake

An irreducible quadratic needs a linear top

Bx+C

, not just a number. And a repeated factor needs both

\dfrac{A}{ax+b}

and

\dfrac{B}{(ax+b)^2}

. Get the template wrong and you lose the mark before you have done any arithmetic.

Common mistake

If the fraction is improper, divide first (§7.1). Set up partial fractions on an improper fraction and it never balances. The polynomial part has no place to go.

Common mistake

When you finish by matching coefficients, compare the highest power first. It usually has the fewest terms. Then work down. Start with the wrong power and you tie yourself up in needless equations.

Now you try

Split $\dfrac{2x^2+7x+8}{(1+x)(2+x)^2}$ into partial fractions. (9709/31 Oct/Nov 2022 Q10)

Your turn— tap to reveal the worked answer (9709/31 Oct/Nov 2022 Q10(a))

There's a repeated factor, so use the template $\dfrac{A}{1+x}+\dfrac{B}{2+x}+\dfrac{C}{(2+x)^2}$ . Multiply up: $2x^2+7x+8=A(2+x)^2+B(1+x)(2+x)+C(1+x)$ .

Put $x=-1$ : $A=3$ ; put $x=-2$ : $C=-2$ ; match $x^2$ : $A+B=2\Rightarrow B=-1$ .

Answer

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