By the end you can
Divide one polynomial by another using long division.
Write the result in the form $P(x) = A(x)Q(x) + R(x)$.
The division theorem
Dividing $23$ by $5$ gives quotient $4$ and remainder $3$, written $23 = 5 \times 4 + 3$, with the remainder smaller than the divisor. Polynomials work the same way, with "smaller" meaning lower degree.
Division theorem
Dividing $P(x)$ by a non-zero polynomial $A(x)$ gives a quotient $Q(x)$ and remainder $R(x)$ with
$$P(x) = A(x)\,Q(x) + R(x), \qquad \deg R < \deg A \text{ (or } R = 0\text{)}.$$
If $R(x) = 0$, then $A(x)$ is a factor of $P(x)$.
Long division
Each round has the same four moves: divide the leading terms, multiply back, subtract, bring down. Stop when what is left has lower degree than the divisor.
Divide $P(x) = 2x^3 - 3x^2 + x - 7$ by $x - 2$.
Subtract: $(2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2$. Bring down: $x^2 + x$.
2x³ ÷ x = 2x². Multiply: 2x²(x − 2) = 2x³ − 4x².
Subtract: $(x^2 + x) - (x^2 - 2x) = 3x$. Bring down: $3x - 7$.
x² ÷ x = x. Multiply: x(x − 2) = x² − 2x.
Subtract: $(3x - 7) - (3x - 6) = -1$. Degree $0$, so stop.
3x ÷ x = 3. Multiply: 3(x − 2) = 3x − 6.
$2x^3 - 3x^2 + x - 7 = (x - 2)(2x^2 + x + 3) - 1$
Common mistake
Write in every missing power with a zero coefficient before you start. Dividing $x^3 - 5$ means dividing $x^3 + 0x^2 + 0x - 5$, otherwise the columns slip.
Example 2: a quadratic divisor ●●● Divide $x^4 + 2x^3 - x + 5$ by $x^2 + 1$.
Subtract $x^2(x^2 + 1) = x^4 + x^2$: left with $2x^3 - x^2 - x + 5$.
Fill the gap: x⁴ + 2x³ + 0x² − x + 5. Then x⁴ ÷ x² = x².
Subtract $2x(x^2 + 1) = 2x^3 + 2x$: left with $-x^2 - 3x + 5$.
2x³ ÷ x² = 2x.
Subtract $-1(x^2 + 1)$: left with $-3x + 6$, degree $1 < 2$, so stop.
−x² ÷ x² = −1.
$x^4 + 2x^3 - x + 5 = (x^2 + 1)(x^2 + 2x - 1) + (-3x + 6)$
Tip
Check your answer by expanding $A(x)Q(x) + R(x)$, or by substituting an easy value such as $x = 0$ or $x = 1$ into both sides.
Division theorem
$P(x) = A(x)Q(x) + R(x)$ with $\deg R < \deg A$.
Remainder when dividing by a quadratic
Degree at most $1$: of the form $ax + b$.
Remainder zero
$A(x)$ is a factor of $P(x)$.
Find the remainder when $2x^3 - 3x^2 + x - 7$ is divided by $x - 2$.
The quotient when $2x^3 - 3x^2 + x - 7$ is divided by $x - 2$ is:
$2x^2 - x + 3$ $2x^2 + x + 3$ $2x^2 + 7x + 15$ $2x^2 + x - 1$ When a polynomial is divided by a cubic, the remainder has degree:
exactly $2$ at most $3$ at most $2$ $0$ The remainder's degree is less than the divisor's.
A degree $5$ polynomial is divided by a cubic. What is the degree of the quotient?
$x^4 + 2x^3 - x + 5 = (x^2 + 1)(x^2 + 2x - 1) + (-3x + c)$. Find $c$.
See Example 2, or put $x = 0$: $5 = -1 + c$.
Before dividing $x^3 - 5$ by $x - 1$, you should write it as:
$x^3 + 0x^2 + 0x - 5$ $x^3 - 5x^2$ $(x - 1)(x^2 - 5)$ $x^2 - 5$ Every power needs its own column.
Divide $x^2 + 5x + 9$ by $x + 2$. Find the remainder.
$x^2 + 5x + 9 = (x + 2)(x + 3) + 3$.
Divide $x^3 - 8$ by $x - 2$. The quotient is $x^2 + 2x + k$. Find $k$.
$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$, a difference of cubes.
If the remainder is $0$ when $P(x)$ is divided by $A(x)$, then:
$P(x) = 0$ $A(x) = 0$ the quotient is $0$ $A(x)$ is a factor of $P(x)$ Try again
New numbers every time. Keep going until you get five in a row.
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