Mathematics 9709 · Pure Mathematics 3

Improper algebraic fractions

A fraction is improper when the top's degree is as big as the bottom's, or bigger. In numbers that's like $\tfrac{11}{4}$ , which you rewrite as $2\tfrac34$ .
Long-divide it into a polynomial + a proper fraction. It is the same long division as §1.4.
That tidy form is exactly what partial fractions (§7.2) needs.

Improper just means $\deg(\text{top})\ge\deg(\text{bottom})$ . Keep long-dividing until the leftover has lower degree than what you're dividing by: $\dfrac{P(x)}{Q(x)}=\text{quotient}+\dfrac{\text{remainder}}{Q(x)}$ . Now the leftover fraction is proper.

Recognise it, then divide

First compare the degrees (top degree 3 over bottom degree 1 gives a degree-2 quotient).
Then divide. Fill any missing power with a $0$ so the columns stay lined up:

$\dfrac{P(x)}{Q(x)}=\text{quotient}+\dfrac{\text{remainder}}{Q(x)},\qquad \deg(\text{remainder})<\deg(Q)$

You can work out the degree of the quotient before you start: $\deg(\text{quotient})=\deg(\text{top})-\deg(\text{bottom})$ . So a degree-2-over-degree-1 fraction must give a quadratic plus a constant over the bottom. That is a quick way to check your answer looks right.

Worked examples

Worked example

Express $\dfrac{4x^3-3}{2x+1}$ as a polynomial plus a proper fraction

Degrees: top is 3, bottom is 1, so the quotient is degree 2. Write the top with every power present: $4x^3+0x^2+0x-3$ .
Long-divide. The quotient comes out $2x^2-x+\tfrac12$ with remainder $-\tfrac72$ : $4x^3-3=(2x+1)\!\left(2x^2-x+\tfrac12\right)-\tfrac72$ .
Divide the remainder back over $2x+1$ :

$\dfrac{4x^3-3}{2x+1}=2x^2-x+\tfrac12-\dfrac{7}{2(2x+1)}$

Answer

2x^2-x+\tfrac12-\dfrac{7}{2(2x+1)}

Worked example

Express $\dfrac{2x^2+5x-11}{x^2+2x-3}$ as polynomial plus proper fraction

Top and bottom have the same degree, so the quotient is just a number.

Both are degree 2, so the quotient is a constant. Divide: $2x^2+5x-11=2(x^2+2x-3)+(x-5)$ , since $2(x^2+2x-3)=2x^2+4x-6$ and the difference is $x-5$ .
The leftover $x-5$ is degree 1, below the divisor's degree 2, so we stop:

Answer

2+\dfrac{x-5}{x^2+2x-3}

Why divide before splitting

Why partial fractions need a proper fraction first

Partial fractions writes a fraction as a sum of proper pieces like $\dfrac{A}{x+3}+\dfrac{B}{x-1}$ . Proper pieces only ever add up to a proper fraction. So if you split an improper fraction straight away, it can never match. The polynomial part has no place to go.

Divide first and you take that polynomial part off. What is left is a proper fraction the split can handle. That is why §7.2's “Case 4” starts with a division.

Where the marks go

Common mistake

Stop dividing the moment the leftover's degree drops below the bottom's. Not sooner, not later. Fill missing powers with a

0

(write

4x^3+0x^2+0x-3

). That keeps the columns lined up so no term slips out of place.

Common mistake

Don't lose the number in the quotient. In the first example the

+\tfrac12

is a real term. Drop it and you ruin the whole answer, and every partial fraction you build on it later.

Now you try

Write $\dfrac{6x^3+7x^2-6}{x-2}$ as a polynomial plus a proper fraction. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Top degree 3, bottom degree 1, so the quotient is degree 2. Dividing gives quotient $6x^2+19x+38$ and remainder $70$ :

Answer

6x^2+19x+38+\dfrac{70}{x-2}

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Improper algebraic fractions

A fraction is improper when the top's degree is as big as the bottom's, or bigger. In numbers that's like $\tfrac{11}{4}$ , which you rewrite as $2\tfrac34$ .
Long-divide it into a polynomial + a proper fraction. It is the same long division as §1.4.
That tidy form is exactly what partial fractions (§7.2) needs.

Recognise it, then divide

First compare the degrees (top degree 3 over bottom degree 1 gives a degree-2 quotient).
Then divide. Fill any missing power with a $0$ so the columns stay lined up:

$\dfrac{P(x)}{Q(x)}=\text{quotient}+\dfrac{\text{remainder}}{Q(x)},\qquad \deg(\text{remainder})<\deg(Q)$

Worked examples

Worked example

Express $\dfrac{4x^3-3}{2x+1}$ as a polynomial plus a proper fraction

Degrees: top is 3, bottom is 1, so the quotient is degree 2. Write the top with every power present: $4x^3+0x^2+0x-3$ .
Long-divide. The quotient comes out $2x^2-x+\tfrac12$ with remainder $-\tfrac72$ : $4x^3-3=(2x+1)\!\left(2x^2-x+\tfrac12\right)-\tfrac72$ .
Divide the remainder back over $2x+1$ :

$\dfrac{4x^3-3}{2x+1}=2x^2-x+\tfrac12-\dfrac{7}{2(2x+1)}$

Answer

2x^2-x+\tfrac12-\dfrac{7}{2(2x+1)}

Worked example

Express $\dfrac{2x^2+5x-11}{x^2+2x-3}$ as polynomial plus proper fraction

Top and bottom have the same degree, so the quotient is just a number.

Both are degree 2, so the quotient is a constant. Divide: $2x^2+5x-11=2(x^2+2x-3)+(x-5)$ , since $2(x^2+2x-3)=2x^2+4x-6$ and the difference is $x-5$ .
The leftover $x-5$ is degree 1, below the divisor's degree 2, so we stop:

Answer

2+\dfrac{x-5}{x^2+2x-3}

Why divide before splitting

Why partial fractions need a proper fraction first

Divide first and you take that polynomial part off. What is left is a proper fraction the split can handle. That is why §7.2's “Case 4” starts with a division.

Where the marks go

Common mistake

Stop dividing the moment the leftover's degree drops below the bottom's. Not sooner, not later. Fill missing powers with a

0

(write

4x^3+0x^2+0x-3

). That keeps the columns lined up so no term slips out of place.

Common mistake

Don't lose the number in the quotient. In the first example the

+\tfrac12

is a real term. Drop it and you ruin the whole answer, and every partial fraction you build on it later.

Now you try

Write $\dfrac{6x^3+7x^2-6}{x-2}$ as a polynomial plus a proper fraction. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Top degree 3, bottom degree 1, so the quotient is degree 2. Dividing gives quotient $6x^2+19x+38$ and remainder $70$ :

Answer

6x^2+19x+38+\dfrac{70}{x-2}