Multiplying and dividing polynomials · Polynomials & algebraic division

A polynomial is a sum of terms in whole-number powers of $x$ , such as $2x^3 - 5x^2 + x - 7$ . The number in front of each power is its coefficient, the term with no $x$ is the constant, and the highest power is the degree. Degree 2 is a quadratic, degree 3 a cubic. Always write a polynomial in descending powers.

2x³ − 5x² + x − 7

leading termcoefficient of x²constant

Highest power 3: degree 3 (a cubic)

Parts of a polynomialLeading term 2x³, coefficient of x² is −5, constant −7, degree 3

Multiplying

Multiply every term of one bracket by every term of the other, then collect like terms:

\begin{aligned} &(x - 2)(x^2 + 3x - 1) \\ &= x^3 + 3x^2 - x - 2x^2 - 6x + 2 \\ &= x^3 + x^2 - 7x + 2 \end{aligned}

×	x²	3x	−1
x	x³	3x²	−x
−2	−2x²	−6x	+2

Every term times every termEach cell is a row term times a column term; add the cells, collecting like powers

Long division

Dividing polynomials works like long division of numbers. Each round has three moves:

Divide the leading term of what’s left by the leading term of the divisor. That gives the next term of the quotient.
Multiply the whole divisor by that term and write the result underneath, lined up by power.
Subtract, then bring down the next term.

Stop when what’s left has a lower degree than the divisor: that is the remainder. Just as $17 = 5 \times 3 + 2$ , the polynomial equals divisor × quotient + remainder.

17 = 5 × 3 + 2

f(x) = divisor × quotient + remainder

The remainder has a lower degree than the divisor

Dividend, divisor, quotient, remainderThe same relationship as for numbers

Try it

Long divisionPress Next step

x + 1 )x³+ 6x²+ 11x+ 6

—quotient so far—remainder

Write the dividend in descending powers. Press Next step.

Step through each example. “Missing term” shows why you write $0x^2$ for a power that isn’t there: without it the columns don’t line up.

Worked example · JAMB 1998

JAMB 1998 · UME · Q15

Divide $2x^3 + 11x^2 + 17x + 6$ by $2x + 1$ .

First round
$2x^3 \div 2x = x^2$ . Then $x^2(2x + 1) = 2x^3 + x^2$ , and subtracting leaves $10x^2 + 17x$ .
Think first. 2x³ ÷ 2x = ? Then multiply (2x + 1) by it.
Second round
$10x^2 \div 2x = 5x$ . Then $5x(2x + 1) = 10x^2 + 5x$ , and subtracting leaves $12x + 6$ .
Think first. 10x² ÷ 2x = ?
Third round
$12x \div 2x = 6$ . Then $6(2x + 1) = 12x + 6$ , which leaves 0.
Think first. 12x ÷ 2x = ?
Answer
$x^2 + 5x + 6$ , with no remainder: option A.
Think first. What is the quotient?

Dividing by a quadratic

The same three moves work for a quadratic divisor. Write in any missing power, so $x^2 + 1$ becomes $x^2 + 0x + 1$ . The remainder can now have an $x$ term, because it only needs a lower degree than the divisor.

Worked example · JAMB 2011

JAMB 2011 · UTME · Q13

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

First round
$x^3 \div x^2 = x$ . Then $x(x^2 + 0x + 1) = x^3 + 0x^2 + x$ , and subtracting leaves $-2x^2 + 2x - 3$ .
Think first. x³ ÷ x² = ? Multiply x² + 0x + 1 by it.
Second round
$-2x^2 \div x^2 = -2$ . Then $-2(x^2 + 0x + 1) = -2x^2 - 2$ , and subtracting leaves $2x - 1$ .
Think first. −2x² ÷ x² = ?
Stop
$2x - 1$ has a lower degree than $x^2 + 1$ , so it is the remainder: option A. (The quotient is $x - 2$ .)
Think first. Can 2x − 1 be divided by x² again?

Your turn

JAMB 1999 · UME · Q16

Polynomials & algebraic division

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

Worked solution (try it first)

Write the missing $x^2$ term: $4x^3 + 0x^2 - 3x + 1$ .
Then $4x^3 \div 2x = 2x^2$ .
Subtract $4x^3 - 2x^2$ to leave $2x^2 - 3x + 1$ .
$2x^2 \div 2x = x$ .
Subtract $2x^2 - x$ to leave $-2x + 1$ .
$-2x \div 2x = -1$ , and $-1(2x - 1) = -2x + 1$ leaves 0.
So the quotient is $2x^2 + x - 1$ , option D.

Report a problem with this question