Polynomial Division

A2 · P3 · 13 min

Dividing one polynomial by another works exactly like long division of whole numbers: you get a quotient and a remainder. In P3 you divide polynomials of degree up to $4$ by a linear or a quadratic divisor. The skill appears on its own ("find the quotient and remainder"), inside factor theorem questions ("hence factorise"), and before integration, where an awkward fraction must first be split into a polynomial plus a proper fraction.

The language of division

Think of $47 \div 5$ . Five goes into $47$ nine times with $2$ left over, so

47 = 5 \times 9 + 2.

The remainder $2$ is smaller than the divisor $5$ ; otherwise another $5$ could have been taken out. Polynomials behave the same way, with "smaller" meaning "lower degree".

Definition

The degree of a polynomial is the highest power of $x$ with a non-zero coefficient. When a polynomial $p(x)$ (the dividend) is divided by a polynomial $d(x)$ (the divisor), there are unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that

p(x) \equiv d(x)\,q(x) + r(x),

where the degree of $r(x)$ is less than the degree of $d(x)$ . The remainder may be zero, in which case $d(x)$ is a factor of $p(x)$ .

Key result

Divisor	Degree of remainder	Form of remainder
linear, e.g. $x - 2$ or $2x + 1$	$0$	a constant $R$
quadratic, e.g. $x^2 + x - 1$	at most $1$	$Rx + S$

The degree of the quotient is (degree of dividend) $-$ (degree of divisor). So a quartic divided by a quadratic gives a quadratic quotient.

The $\equiv$ sign means the two sides are equal for every value of $x$ : it is an identity, not an equation to be solved. That is what lets you compare coefficients or substitute any convenient value of $x$ .

Long division

The procedure is a loop of four steps, repeated until what is left has lower degree than the divisor.

Polynomial long division

Write the dividend in descending powers of $x$ , inserting $0x^k$ for any missing power.
Divide the leading term of what is left by the leading term of the divisor. This is the next term of the quotient.
Multiply the whole divisor by that term.
Subtract the result from what is left. The leading term cancels.
Repeat steps 2 to 4 until the degree of what is left is less than the degree of the divisor. What is left is the remainder.
Check by expanding $d(x)q(x) + r(x)$ or by substituting a value such as $x = 1$ .

Dividing a cubic by a linear divisor

Find the quotient and remainder when $2x^3 - 5x^2 + x + 6$ is divided by $x - 2$ .

Solution

Work through the loop, recording each stage.

Step	Divide	Multiply back	Subtract
1	$2x^3 \div x = 2x^2$	$2x^2(x - 2) = 2x^3 - 4x^2$	$(2x^3 - 5x^2 + x + 6) - (2x^3 - 4x^2) = -x^2 + x + 6$
2	$-x^2 \div x = -x$	$-x(x - 2) = -x^2 + 2x$	$(-x^2 + x + 6) - (-x^2 + 2x) = -x + 6$
3	$-x \div x = -1$	$-1(x - 2) = -x + 2$	$(-x + 6) - (-x + 2) = 4$

What is left, $4$ , has degree $0$ , less than the degree of $x - 2$ , so stop.

Quotient $2x^2 - x - 1$ , remainder $4$ :

2x^3 - 5x^2 + x + 6 \equiv (x - 2)(2x^2 - x - 1) + 4.

Check at $x = 1$ : left side $2 - 5 + 1 + 6 = 4$ ; right side $(-1)(0) + 4 = 4$ .

Watch out

The subtraction step is where most errors happen. Subtracting $(-x^2 + 2x)$ from $(-x^2 + x)$ gives $-x$ , because $x - 2x = -x$ . Put brackets round the expression being subtracted and change every sign inside.

Missing powers

Divide $x^4 - 3x^2 + 5$ by $x + 1$ .

Solution

There is no $x^3$ term and no $x$ term, so write the dividend as $x^4 + 0x^3 - 3x^2 + 0x + 5$ . Without the placeholders the columns slip and the answer is wrong.

Step	Divide	Multiply back	Subtract
1	$x^4 \div x = x^3$	$x^4 + x^3$	$-x^3 - 3x^2$
2	$-x^3 \div x = -x^2$	$-x^3 - x^2$	$-2x^2 + 0x$
3	$-2x^2 \div x = -2x$	$-2x^2 - 2x$	$2x + 5$
4	$2x \div x = 2$	$2x + 2$	$3$

Quotient $x^3 - x^2 - 2x + 2$ , remainder $3$ .

Check at $x = 1$ : left $1 - 3 + 5 = 3$ ; right $2 \times (1 - 1 - 2 + 2) + 3 = 0 + 3 = 3$ .

Dividing by $(ax + b)$ when $a \ne 1$

Nothing changes in the method. The quotient may pick up fractions if the leading coefficients do not divide neatly, but the remainder is still a constant.

A non-monic linear divisor

Find the quotient and remainder when $6x^3 + 5x^2 - 8x + 3$ is divided by $2x - 1$ .

Solution

Step	Divide	Multiply back	Subtract
1	$6x^3 \div 2x = 3x^2$	$3x^2(2x - 1) = 6x^3 - 3x^2$	$8x^2 - 8x$
2	$8x^2 \div 2x = 4x$	$4x(2x - 1) = 8x^2 - 4x$	$-4x + 3$
3	$-4x \div 2x = -2$	$-2(2x - 1) = -4x + 2$	$1$

Quotient $3x^2 + 4x - 2$ , remainder $1$ :

6x^3 + 5x^2 - 8x + 3 \equiv (2x - 1)(3x^2 + 4x - 2) + 1.

Check at $x = \tfrac{1}{2}$ : $6 \cdot \tfrac{1}{8} + 5 \cdot \tfrac{1}{4} - 4 + 3 = \tfrac{3}{4} + \tfrac{5}{4} - 1 = 1$ . This matches the remainder, as the remainder theorem predicts.

Comparing coefficients (division by inspection)

Instead of long division you can write the identity with unknown coefficients and match powers of $x$ . It is often faster, and it is the standard way to find the quadratic factor once you know a linear factor.

Dividing by comparing coefficients

Write $p(x) \equiv d(x)\,q(x) + r(x)$ with $q(x)$ and $r(x)$ of the correct degrees and unknown coefficients.
Match the highest power first: it fixes the leading coefficient of $q$ .
Match the constant term next if it involves only one unknown.
Work through the remaining powers one at a time.

Dividing a quartic by a quadratic

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

Solution

The quotient is a quadratic and the remainder is linear:

x^4 + 3x^3 + 0x^2 - 2x + 1 \equiv (x^2 + x - 1)(x^2 + px + q) + (rx + s).

Expand the product: $x^4 + (p + 1)x^3 + (q + p - 1)x^2 + (q - p)x - q$ . Now compare:

$x^3$ : $p + 1 = 3$ , so $p = 2$ .
$x^2$ : $q + p - 1 = 0$ , so $q = -1$ .
$x^1$ : $(q - p) + r = -2$ , so $-3 + r = -2$ , giving $r = 1$ .
$x^0$ : $-q + s = 1$ , so $1 + s = 1$ , giving $s = 0$ .

Quotient $x^2 + 2x - 1$ , remainder $x$ .

Long division gives the same result:

Step	Divide	Multiply back	Subtract
1	$x^4 \div x^2 = x^2$	$x^4 + x^3 - x^2$	$2x^3 + x^2 - 2x$
2	$2x^3 \div x^2 = 2x$	$2x^3 + 2x^2 - 2x$	$-x^2 + 0x + 1$
3	$-x^2 \div x^2 = -1$	$-x^2 - x + 1$	$x$

What is left, $x$ , has degree $1 < 2$ , so it is the remainder.

Tip

For a linear divisor $x - a$ you may also use synthetic division: write the coefficients in a row, bring the first one down, multiply by $a$ , add to the next coefficient, and repeat. For $2x^3 - 5x^2 + x + 6$ divided by $x - 2$ :

	$2$	$-5$	$1$	$6$
$\times 2$		$4$	$-2$	$-2$
sum	$2$	$-1$	$-1$	$4$

The bottom row gives the quotient coefficients $2, -1, -1$ and the remainder $4$ . It is not required by the syllabus, and you should label it clearly if you use it, but it is a fast check.

Division with unknown coefficients

When the polynomial contains unknown constants, divide in the ordinary way, carrying the letters along. The remainder comes out in terms of the unknowns, and the information you are given about the remainder produces equations.

Finding constants from a given remainder

When $x^4 + 2x^3 + ax^2 + bx + 3$ is divided by $x^2 + 1$ , the remainder is $2x - 1$ . Find the values of $a$ and $b$ and state the quotient.

Solution

Long division, with the dividend $x^4 + 2x^3 + ax^2 + bx + 3$ :

Step	Divide	Multiply back	Subtract
1	$x^4 \div x^2 = x^2$	$x^4 + x^2$	$2x^3 + (a - 1)x^2 + bx + 3$
2	$2x^3 \div x^2 = 2x$	$2x^3 + 2x$	$(a - 1)x^2 + (b - 2)x + 3$
3	$(a - 1)x^2 \div x^2 = a - 1$	$(a - 1)x^2 + (a - 1)$	$(b - 2)x + (4 - a)$

So the remainder is $(b - 2)x + (4 - a)$ . Matching with $2x - 1$ :

b - 2 = 2 \Rightarrow b = 4, \qquad 4 - a = -1 \Rightarrow a = 5.

The quotient is $x^2 + 2x + (a - 1) = x^2 + 2x + 4$ .

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

Using division to rewrite a fraction

If the top of a fraction has degree at least that of the bottom, division rewrites it as a polynomial plus a proper fraction (one whose numerator has lower degree than its denominator):

\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}.

This is the first step before partial fractions or integration whenever the fraction is not already proper.

Preparing a fraction for integration

Express $\dfrac{2x^3 + 3x^2 - 4x + 1}{x^2 - 1}$ in the form $ax + b + \dfrac{cx + d}{x^2 - 1}$ , where $a$ , $b$ , $c$ and $d$ are constants.

Solution

Divide $2x^3 + 3x^2 - 4x + 1$ by $x^2 + 0x - 1$ :

Step	Divide	Multiply back	Subtract
1	$2x^3 \div x^2 = 2x$	$2x^3 - 2x$	$3x^2 - 2x + 1$
2	$3x^2 \div x^2 = 3$	$3x^2 - 3$	$-2x + 4$

Quotient $2x + 3$ , remainder $-2x + 4$ . So

\frac{2x^3 + 3x^2 - 4x + 1}{x^2 - 1} = 2x + 3 + \frac{4 - 2x}{x^2 - 1},

with $a = 2$ , $b = 3$ , $c = -2$ , $d = 4$ .

Check at $x = 0$ : left $\dfrac{1}{-1} = -1$ ; right $3 + \dfrac{4}{-1} = -1$ .

Common mistakes

Watch out

Forgetting placeholder zeros. Dividing $x^4 - 3x^2 + 5$ without writing $0x^3$ and $0x$ misaligns every later step. Always write the dividend in full.

Watch out

Stopping too early or too late. Stop when the degree of what is left is less than the degree of the divisor. For a quadratic divisor, a remainder like $3x + 2$ is fine, but $x^2 + 1$ is not finished.

Watch out

Sign slips in the subtraction. Bracket what you subtract and change every sign. Check the final answer by substituting $x = 1$ or $x = 0$ into the identity.

Watch out

Confusing the quotient with the remainder. "Find the quotient" wants $q(x)$ ; "find the remainder" wants $r(x)$ . When asked for both, label them.

Exam tip

Questions typically say "Find the quotient and remainder when $p(x)$ is divided by $x^2 + x - 1$ ". Long division or comparing coefficients are both fully acceptable; the method mark needs a clear sequence of working, not just answers.
Write the final statement in the identity form $p(x) \equiv d(x)q(x) + r(x)$ , or clearly state "quotient $\ldots$ , remainder $\ldots$ ".
A remainder of zero means the divisor is a factor. Many questions end "hence show that $p(x) = 0$ has only one real root" or similar: divide, then examine the discriminant of the quadratic quotient (see the factor theorem).
The syllabus caps the dividend at degree $4$ and the divisor at degree $2$ , so the working is never longer than four rows.

Summary

$p(x) \equiv d(x)q(x) + r(x)$ with $\deg r < \deg d$ ; it is an identity, true for all $x$ .
Linear divisor: constant remainder. Quadratic divisor: remainder $Rx + S$ .
Long division: divide leading terms, multiply back, subtract, repeat.
Always insert $0x^k$ for missing powers.
Comparing coefficients: write the identity with unknowns, match from the highest power down.
With unknown constants in the dividend, carry them through and match the remainder you are given.
To rewrite an improper fraction, divide: $\dfrac{p}{d} = q + \dfrac{r}{d}$ .

Practice

Question

Find the quotient and remainder when $x^3 - 7x + 6$ is divided by $x + 3$ .
Find the quotient and remainder when $3x^3 - 2x^2 + 4$ is divided by $x - 3$ .
Divide $x^3 + 8$ by $x + 2$ .
Find the quotient and remainder when $6x^3 - 7x^2 + x + 2$ is divided by $3x + 1$ .
Find the quotient and remainder when $2x^4 - x^3 + 3x - 5$ is divided by $x^2 + 2$ .
Find the quotient and remainder when $x^4 - 2x^3 + 3x^2 - x + 4$ is divided by $x^2 - x + 2$ .
The polynomial $x^4 + x^3 + px + q$ is divided by $x^2 + x + 1$ . The remainder is $2x + 5$ . Find $p$ and $q$ .
Express $\dfrac{x^3 + 2x^2 - 1}{x^2 + 1}$ in the form $ax + b + \dfrac{cx + d}{x^2 + 1}$ .

Answers

Write $x^3 + 0x^2 - 7x + 6$ . Steps: $x^2$ (subtract $x^3 + 3x^2$ , leaving $-3x^2 - 7x$ ); $-3x$ (subtract $-3x^2 - 9x$ , leaving $2x + 6$ ); $2$ (subtract $2x + 6$ , leaving $0$ ). Quotient $x^2 - 3x + 2$ , remainder $0$ , so $x + 3$ is a factor.
Write $3x^3 - 2x^2 + 0x + 4$ . Steps: $3x^2$ (leaving $7x^2 + 0x$ ); $7x$ (leaving $21x + 4$ ); $21$ (leaving $67$ ). Quotient $3x^2 + 7x + 21$ , remainder $67$ . Check: $3(27) - 2(9) + 4 = 67$ .
Write $x^3 + 0x^2 + 0x + 8$ . Quotient $x^2 - 2x + 4$ , remainder $0$ , so $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ .
Steps: $6x^3 \div 3x = 2x^2$ , subtract $6x^3 + 2x^2$ to leave $-9x^2 + x$ ; $-9x^2 \div 3x = -3x$ , subtract $-9x^2 - 3x$ to leave $4x + 2$ ; $4x \div 3x = \tfrac{4}{3}$ , subtract $4x + \tfrac{4}{3}$ to leave $\tfrac{2}{3}$ . Quotient $2x^2 - 3x + \tfrac{4}{3}$ , remainder $\tfrac{2}{3}$ . Check: $f\left(-\tfrac{1}{3}\right) = -\tfrac{6}{27} - \tfrac{7}{9} - \tfrac{1}{3} + 2 = \tfrac{2}{3}$ .
Write $2x^4 - x^3 + 0x^2 + 3x - 5$ . Steps: $2x^2$ (subtract $2x^4 + 4x^2$ , leaving $-x^3 - 4x^2 + 3x$ ); $-x$ (subtract $-x^3 - 2x$ , leaving $-4x^2 + 5x - 5$ ); $-4$ (subtract $-4x^2 - 8$ , leaving $5x + 3$ ). Quotient $2x^2 - x - 4$ , remainder $5x + 3$ .
Steps: $x^2$ (subtract $x^4 - x^3 + 2x^2$ , leaving $-x^3 + x^2 - x$ ); $-x$ (subtract $-x^3 + x^2 - 2x$ , leaving $x + 4$ ). Degree $1 < 2$ , so stop. Quotient $x^2 - x$ , remainder $x + 4$ .
Divide $x^4 + x^3 + 0x^2 + px + q$ by $x^2 + x + 1$ . Steps: $x^2$ (subtract $x^4 + x^3 + x^2$ , leaving $-x^2 + px + q$ ); $-1$ (subtract $-x^2 - x - 1$ , leaving $(p + 1)x + (q + 1)$ ). So $p + 1 = 2$ and $q + 1 = 5$ : $p = 1$ , $q = 4$ . Quotient $x^2 - 1$ .
Divide $x^3 + 2x^2 + 0x - 1$ by $x^2 + 1$ : $x$ (subtract $x^3 + x$ , leaving $2x^2 - x - 1$ ); $2$ (subtract $2x^2 + 2$ , leaving $-x - 3$ ). So $\dfrac{x^3 + 2x^2 - 1}{x^2 + 1} = x + 2 + \dfrac{-x - 3}{x^2 + 1}$ : $a = 1$ , $b = 2$ , $c = -1$ , $d = -3$ .

The language of division

Long division

Dividing by (ax+b)(ax + b)(ax+b) when a≠1a \ne 1a=1

Comparing coefficients (division by inspection)

Division with unknown coefficients

Using division to rewrite a fraction

Common mistakes

Summary

Practice

Console

Dividing by $(ax + b)$ when $a \ne 1$