From Zero to Engineer

Multiplication and division of algebraic expressions

Multiplying polynomials in columns with terms of the same degree lined up, and dividing a polynomial by a binomial step by step, with the result checked by multiplying back.

Polynomials are multiplied and divided just like multi-digit numbers - in columns. The only difference is that instead of columns for units, tens and hundreds we have columns for successive powers of x. That is why a missing power is always written in with a zero: without it the terms stop lining up and the calculation drifts. The result of a division can be checked immediately by multiplying the quotient by the divisor.

Multiplication and division of algebraic expressions

Polynomials are multiplied just like numbers in columns: each term of the lower polynomial multiplies the whole upper one, and the partial results are written under one another with terms of the same degree lined up.

(2x + 5)(x^{2} + 3x + 4) = 2x^{3} + 11x^{2} + 23x + 20

x^{2} + 3x + 4

2x + 5

$2x^{3}$	$+ 6x^{2}$	$+ 8x$
	$5x^{2}$	$+ 15x$	$+ 20$
$2x^{3}$	$+ 11x^{2}$	$+ 23x$	$+ 20$

Missing powers are written in with a zero coefficient, which keeps the columns lined up.

(2x + 6)(4x^{3} - 5x - 7) = (2x + 6)(4x^{3} + 0x^{2} - 5x - 7)

4x^{3} + 0x^{2} - 5x - 7

2x + 6

$8x^{4}$	$+ 0x^{3}$	$- 10x^{2}$	$- 14x$
	$24x^{3}$	$+ 0x^{2}$	$- 30x$	$- 42$
$8x^{4}$	$+ 24x^{3}$	$- 10x^{2}$	$- 44x$	$- 42$

(2x + 6)(4x^{3} - 5x - 7) = 8x^{4} + 24x^{3} - 10x^{2} - 44x - 42

(3x - 5)(2x^{3} - 4x^{2} + 8) = (3x - 5)(2x^{3} - 4x^{2} + 0x + 8)

2x^{3} - 4x^{2} + 0x + 8

3x - 5

$6x^{4}$	$- 12x^{3}$	$+ 0x^{2}$	$+ 24x$
	$- 10x^{3}$	$+ 20x^{2}$	$+ 0x$	$- 40$
$6x^{4}$	$- 22x^{3}$	$+ 20x^{2}$	$+ 24x$	$- 40$

(3x - 5)(2x^{3} - 4x^{2} + 8) = 6x^{4} - 22x^{3} + 20x^{2} + 24x - 40

Division of polynomials

We divide as with numbers: take the highest term of the dividend, divide it by the highest term of the divisor, multiply the result by the whole divisor and subtract. Repeat until nothing is left. The answer can always be checked by multiplying back.

(12x^{3} - 2x^{2} - 3x + 28) : (3x + 4) = 4x^{2} - 6x + 7

$12x^{3}$	$+ 16x^{2}$
0	$- 18x^{2}$	$- 3x$
	$- 18x^{2}$	$- 24x$
	0	$21x$	$+ 28$
		$21x$	$+ 28$
			$0$

(3x + 4)(4x^{2} - 6x + 7) = 12x^{3} - 2x^{2} - 3x + 28

(4x^{3} + 0x^{2} + 13x + 33) : (2x + 3) = 2x^{2} - 3x + 11

$4x^{3}$	$+ 6x^{2}$
0	$- 6x^{2}$	$+ 13x$
	$- 6x^{2}$	$- 9x$
	0	$22x$	$+ 33$
		$22x$	$+ 33$
			$0$

(2x + 3)(2x^{2} - 3x + 11) = 4x^{3} + 0x^{2} + 13x + 33

(6x^{3} - 7x^{2} + 0x + 1) : (3x + 1) = 2x^{2} - 3x + 1

$6x^{3}$	$+ 2x^{2}$
0	$- 9x^{2}$	$+ 0x$
	$- 9x^{2}$	$- 3x$
	0	$3x$	$+ 1$
		$3x$	$+ 1$
			$0$

(3x + 1)(2x^{2} - 3x + 1) = 6x^{3} - 7x^{2} + 0x + 1

Review exercises

Multiply and simplify the results.

\text{a)}\quad (8x - 4)(4x^{2} - 3x + 2)

4x^{2} - 3x + 2

8x - 4

$32x^{3}$	$- 24x^{2}$	$+ 16x$
	$- 16x^{2}$	$+ 12x$	$- 8$
$32x^{3}$	$- 40x^{2}$	$+ 28x$	$- 8$

(8x - 4)(4x^{2} - 3x + 2) = 32x^{3} - 40x^{2} + 28x - 8

\text{b)}\quad (2x + 3)(5x^{3} + 3x - 4) = (2x + 3)(5x^{3} + 0x^{2} + 3x - 4)

5x^{3} + 0x^{2} + 3x - 4

2x + 3

$10x^{4}$	$+ 0x^{3}$	$+ 6x^{2}$	$- 8x$
	$15x^{3}$	$+ 0x^{2}$	$+ 9x$	$- 12$
$10x^{4}$	$+ 15x^{3}$	$+ 6x^{2}$	$+ x$	$- 12$

(2x + 3)(5x^{3} + 3x - 4) = 10x^{4} + 15x^{3} + 6x^{2} + x - 12

Perform the division.

(x^{2} + 5x - 6) : (x - 1) = x + 6

$x^{2}$	$- x$
0	$6x$	$- 6$
	$6x$	$- 6$
		$0$

(x - 1)(x + 6) = x^{2} + 5x - 6

(x^{2} - x - 2) : (x + 1) = x - 2

$x^{2}$	$+ x$
0	$- 2x$	$- 2$
	$- 2x$	$- 2$
		$0$

(x + 1)(x - 2) = x^{2} - x - 2

(12x^{3} - 11x^{2} + 0x - 25) : (3x - 5) = 4x^{2} + 3x + 5

$12x^{3}$	$- 20x^{2}$
0	$9x^{2}$	$+ 0x$
	$9x^{2}$	$- 15x$
	0	$15x$	$- 25$
		$15x$	$- 25$
			$0$

(3x - 5)(4x^{2} + 3x + 5) = 12x^{3} - 11x^{2} + 0x - 25

Frequently asked questions

How do you multiply two polynomials?

Multiply every term of one by every term of the other, then combine like terms. In columns you do it row by row: the whole upper polynomial times the first term of the lower one, then times the second, and finally add the columns.

Why write in a term like 0x²?

So that every power has its own column. The polynomial 4x³ − 5x − 7 has no x² term, so we write it as 4x³ + 0x² − 5x − 7 and then terms of the same degree stand exactly under one another.

How do you divide a polynomial by a binomial?

Divide the highest term of the dividend by the highest term of the divisor - that is the first term of the quotient. Multiply it by the whole divisor, subtract from the dividend and repeat with what is left until you reach zero.

How do you check the result of a division?

Multiply the quotient by the divisor - you must get the dividend back. For (12x³ − 2x² − 3x + 28) : (3x + 4) = 4x² − 6x + 7 you check that (3x + 4)(4x² − 6x + 7) gives back 12x³ − 2x² − 3x + 28.

← Previous

Logarithmic equations and rearranging formulas

Algebraic fractions: addition, multiplication and division