Long division and the division algorithm

You already know that dividing $9$ by $4$ gives quotient $2$ and remainder $1$ , because $9 = 4 \times 2 + 1$ . Polynomials divide in exactly the same way, and the method is the long division you learnt for numbers. This lesson shows how to set out the working, how to check it, and the rule behind it, called the division algorithm. Everything that follows in this chapter, the remainder theorem, the factor theorem and factorisation, rests on it.

The division algorithm

If $f(x)$ and $g(x)$ are two polynomials, with $g(x)$ not the zero polynomial, then there exist unique polynomials $q(x)$ and $r(x)$ such that

$f(x) = g(x)\,q(x) + r(x)$

where either $r(x) = 0$ or $\deg r(x) < \deg g(x)$ .

Here $f(x)$ is the dividend, $g(x)$ the divisor, $q(x)$ the quotient and $r(x)$ the remainder. ("deg" is short for degree.) Compare it with numbers: $9 = 4 \times 2 + 1$ , with the remainder $1$ smaller than the divisor $4$ . For polynomials, "smaller" means "of lower degree", so we keep dividing until what is left has a lower degree than the divisor.

The method

Before you start, write both polynomials in descending powers, and fill any missing power with a zero term. Then repeat four moves:

Divide the leading term of what is left by the leading term of the divisor. This gives the next term of the quotient.
Multiply the whole divisor by that term.
Subtract the product from what is left. (Change every sign of the product and add.)
Bring down the next term, and go back to step 1.

Stop when what is left has a lower degree than the divisor. That is the remainder.

Example 1

Find the quotient and remainder when $f(x) = 2x^4 - 6x^3 + 2x^2 - x + 2$ is divided by $x + 2$ .

Each row of the table is one round of the four moves.

Round	Divide	Multiply by $(x + 2)$	Left after subtracting
1	$2x^4 \div x = 2x^3$	$2x^4 + 4x^3$	$-10x^3 + 2x^2 - x + 2$
2	$-10x^3 \div x = -10x^2$	$-10x^3 - 20x^2$	$22x^2 - x + 2$
3	$22x^2 \div x = 22x$	$22x^2 + 44x$	$-45x + 2$
4	$-45x \div x = -45$	$-45x - 90$	$92$

Look closely at round 2: subtracting $-20x^2$ from $2x^2$ gives $2x^2 + 20x^2 = 22x^2$ . Subtracting a negative is where most mistakes happen.

The terms in the "Divide" column make the quotient:

$q(x) = 2x^3 - 10x^2 + 22x - 45, \qquad r(x) = 92$

Check. $(x + 2)(2x^3 - 10x^2 + 22x - 45) = 2x^4 - 6x^3 + 2x^2 - x - 90$ , and adding the remainder $92$ gives back $2x^4 - 6x^3 + 2x^2 - x + 2$ . So $f(x) = g(x)q(x) + r(x)$ holds.

Example 2

Divide $x^3 - 6x^2 + 11x - 6$ by $x^2 - 3x + 2$ .

Round	Divide	Multiply by $(x^2 - 3x + 2)$	Left after subtracting
1	$x^3 \div x^2 = x$	$x^3 - 3x^2 + 2x$	$-3x^2 + 9x - 6$
2	$-3x^2 \div x^2 = -3$	$-3x^2 + 9x - 6$	$0$

Quotient $= x - 3$ , remainder $= 0$ . A zero remainder means $x^2 - 3x + 2$ divides the dividend exactly:

$x^3 - 6x^2 + 11x - 6 = (x^2 - 3x + 2)(x - 3)$

Example 3

Divide $4x^4 - 2x^3 - 6x^2 + x - 5$ by $2x^2 + x - 1$ .

Round	Divide	Multiply by $(2x^2 + x - 1)$	Left after subtracting
1	$4x^4 \div 2x^2 = 2x^2$	$4x^4 + 2x^3 - 2x^2$	$-4x^3 - 4x^2 + x - 5$
2	$-4x^3 \div 2x^2 = -2x$	$-4x^3 - 2x^2 + 2x$	$-2x^2 - x - 5$
3	$-2x^2 \div 2x^2 = -1$	$-2x^2 - x + 1$	$-6$

Quotient $= 2x^2 - 2x - 1$ , remainder $= -6$ . The remainder $-6$ has degree $0$ , lower than the degree $2$ of the divisor, so we stop.

Example 4

Divide $2x^3 + x - 12$ by $3 - 2x$ .

First tidy both polynomials. Write the divisor in descending powers as $-2x + 3$ , and give the dividend its missing $x^2$ term: $2x^3 + 0x^2 + x - 12$ . The leading term of the divisor is now $-2x$ , and the signs need care.

Round	Divide	Multiply by $(-2x + 3)$	Left after subtracting
1	$2x^3 \div (-2x) = -x^2$	$2x^3 - 3x^2$	$3x^2 + x - 12$
2	$\displaystyle 3x^2 \div (-2x) = -\frac{3}{2}x$	$\displaystyle 3x^2 - \frac{9}{2}x$	$\displaystyle \frac{11}{2}x - 12$
3	$\displaystyle \frac{11}{2}x \div (-2x) = -\frac{11}{4}$	$\displaystyle \frac{11}{2}x - \frac{33}{4}$	$\displaystyle -\frac{15}{4}$

In round 2, $\displaystyle x - \left(-\frac{9}{2}x\right) = \frac{2}{2}x + \frac{9}{2}x = \frac{11}{2}x$ . In round 3, $\displaystyle -12 + \frac{33}{4} = \frac{-48 + 33}{4} = -\frac{15}{4}$ .

$\displaystyle q(x) = -x^2 - \frac{3}{2}x - \frac{11}{4}, \qquad r(x) = -\frac{15}{4}$

Fractions in the quotient are perfectly normal when the leading coefficient of the divisor does not divide the dividend's coefficients.

Using the algorithm backwards

The equation $f(x) = g(x)q(x) + r(x)$ also lets us rebuild a dividend from the other three parts.

Example 5

A polynomial divided by $x - 1$ gives quotient $x^2 + 4$ and remainder $-3$ . Find the polynomial.

$\begin{aligned} f(x) &= (x - 1)(x^2 + 4) + (-3) \\ &= x^3 + 4x - x^2 - 4 - 3 \\ &= x^3 - x^2 + 4x - 7 \end{aligned}$

Practice

Divide $3x^2 + 5x - 1$ by $x + 2$ and verify the division algorithm.
Divide $2x^2 + 3x + 1$ by $x - 2$ .
Divide $3x^3 + x^2 - 20x + 12$ by $3x - 2$ .
Divide $x^3 + 2x^2 - 5x - 6$ by $x^2 - x - 2$ .
Divide $3x^3 + x^2 + 2x + 5$ by $x^2 + 2x + 1$ .
Divide $6x^3 + 5x^2 - 4$ by $2x - 1$ .
Find the quotient when $\displaystyle \frac{1}{8}a^3 - \frac{9}{4}a^2x + \frac{27}{2}ax^2 - 27x^3$ is divided by $\displaystyle \frac{1}{2}a - 3x$ .
A polynomial divided by $x - 3$ gives quotient $x^2 + 2$ and remainder $5$ . Find the polynomial.

Answers

Show answers

Quotient $3x - 1$ , remainder $1$ . Check: $(x + 2)(3x - 1) + 1 = 3x^2 + 5x - 2 + 1 = 3x^2 + 5x - 1$ .
Quotient $2x + 7$ , remainder $15$ .
Quotient $x^2 + x - 6$ , remainder $0$ . (Rounds: $x^2$ , then $x$ , then $-6$ .)
Quotient $x + 3$ , remainder $0$ .
Quotient $3x - 5$ , remainder $9x + 10$ . After the first round, $3x(x^2 + 2x + 1) = 3x^3 + 6x^2 + 3x$ leaves $-5x^2 - x + 5$ ; then $-5(x^2 + 2x + 1)$ leaves $9x + 10$ , which has degree $1 < 2$ .
Write the dividend as $6x^3 + 5x^2 + 0x - 4$ . Quotient $3x^2 + 4x + 2$ , remainder $-2$ .
$\displaystyle \frac{1}{4}a^2 - 3ax + 9x^2$ , with remainder $0$ . (The dividend is $\displaystyle \left(\frac{1}{2}a - 3x\right)^3$ , so the quotient is $\displaystyle \left(\frac{1}{2}a - 3x\right)^2$ .)
$(x - 3)(x^2 + 2) + 5 = x^3 - 3x^2 + 2x - 6 + 5 = x^3 - 3x^2 + 2x - 1$

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Long division and the division algorithm

The divi­sion algo­rithm#

The method#

Exam­ple 1#

Exam­ple 2#

Exam­ple 3#

Exam­ple 4#

Using the algo­rithm back­wards#

Exam­ple 5#

Prac­tice#

Answers#