Dividing Polynomials

You already know how to add, subtract and multiply polynomials. Dividing them works just like long division with numbers: you get a quotient and a remainder. Polynomial division is the engine behind the remainder theorem, the factor theorem, and factoring cubics and quartics, so it’s worth getting fast and accurate at it.

Key ideas

The division statement

When you divide $47$ by $5$, you get $9$ remainder $2$, and you can check it by writing $47 = 5 \times 9 + 2$. Polynomials work the same way. Dividing a polynomial $P(x)$ by a divisor $D(x)$ gives a quotient $Q(x)$ and a remainder $R$ with

$$P(x) = D(x)\,Q(x) + R$$

Part	Name	Example: dividing by $x - 1$
$P(x)$	dividend	$x^3 + 4x^2 - 3x + 5$
$D(x)$	divisor	$x - 1$
$Q(x)$	quotient	$x^2 + 5x + 2$
$R$	remainder	$7$

You stop dividing when the remainder has a lower degree than the divisor. For a linear divisor like $x - 1$, that means the remainder is a constant. (For a quadratic divisor, the remainder can be linear, like $-3x + 4$.)

The division statement is also your check: multiply $D(x)Q(x)$, add $R$, and you should get $P(x)$ back.

Long division

Write both polynomials in descending powers of $x$. If a power is missing, put in a placeholder with coefficient $0$, so the columns line up. Then repeat four steps:

1. Divide the leading term of what’s left by the leading term of the divisor. That’s the next term of the quotient.
2. Multiply that term by the whole divisor.
3. Subtract (change every sign, then add).
4. Bring down the next term.

Long division of $2x^3 + 3x^2 - 4$ by $x + 2$. The placeholder $0x$ keeps the columns lined up.

Synthetic division

When the divisor has the form $x - a$, synthetic division is a shortcut that uses only the coefficients. For $(x^3 + 4x^2 - 3x + 5) \div (x - 1)$, use $a = 1$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 1 & 4 & -3 & 5 \\ & & 1 & 5 & 2 \\ \hline & 1 & 5 & 2 & \boxed{7} \end{array}$$

1. Write $a$ on the left and the coefficients of $P(x)$ across the top (with $0$ for any missing power).
2. Bring the first coefficient straight down.
3. Multiply it by $a$, write the result under the next coefficient, and add.
4. Repeat to the end. The last number is the remainder. The others are the coefficients of the quotient, which is one degree lower than $P(x)$.

Here the bottom row $1,\ 5,\ 2 \mid 7$ means $Q(x) = x^2 + 5x + 2$ and $R = 7$.

Careful with the sign of $a$: for the divisor $x + 2$, write it as $x - (-2)$, so $a = -2$.

Writing the answer as a quotient plus a fraction

Dividing both sides of $P(x) = (x - a)Q(x) + R$ by $x - a$ gives another useful form:

$$\frac{P(x)}{x - a} = Q(x) + \frac{R}{x - a}, \qquad x \ne a$$

For example, $\dfrac{x^3 + 4x^2 - 3x + 5}{x - 1} = x^2 + 5x + 2 + \dfrac{7}{x - 1}$. This form is handy later, when you graph rational functions.

Worked examples

Example 1: Long division

Divide $x^3 + 4x^2 - 3x + 5$ by $x - 1$. Write the division statement and check it.

Solution. Work one term of the quotient at a time.

• $x^3 \div x = x^2$. Multiply: $x^2(x - 1) = x^3 - x^2$. Subtract: $(x^3 + 4x^2) - (x^3 - x^2) = 5x^2$. Bring down $-3x$ to get $5x^2 - 3x$.
• $5x^2 \div x = 5x$. Multiply: $5x(x - 1) = 5x^2 - 5x$. Subtract: $(5x^2 - 3x) - (5x^2 - 5x) = 2x$. Bring down $+5$ to get $2x + 5$.
• $2x \div x = 2$. Multiply: $2(x - 1) = 2x - 2$. Subtract: $(2x + 5) - (2x - 2) = 7$.

The quotient is $x^2 + 5x + 2$ and the remainder is $7$:

$$x^3 + 4x^2 - 3x + 5 = (x - 1)(x^2 + 5x + 2) + 7$$

Check: $(x - 1)(x^2 + 5x + 2) = x^3 + 5x^2 + 2x - x^2 - 5x - 2 = x^3 + 4x^2 - 3x - 2$. Adding $7$ gives $x^3 + 4x^2 - 3x + 5$. ✓

Example 2: A missing term

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

Solution. There’s no $x$ term, so write the dividend as $2x^3 + 3x^2 + 0x - 4$. The figure in Key ideas shows the full layout:

• $2x^3 \div x = 2x^2$. Subtract $2x^2(x + 2) = 2x^3 + 4x^2$ to get $-x^2$, then bring down $0x$.
• $-x^2 \div x = -x$. Subtract $-x(x + 2) = -x^2 - 2x$ to get $2x$, then bring down $-4$.
• $2x \div x = 2$. Subtract $2(x + 2) = 2x + 4$ to get $-8$.

$$2x^3 + 3x^2 - 4 = (x + 2)(2x^2 - x + 2) - 8$$

Check with $x = 0$: the left side is $-4$, and the right side is $(2)(2) - 8 = -4$. ✓

Example 3: Synthetic division

Use synthetic division to divide $2x^4 - 5x^3 + 7x - 3$ by $x - 2$. Write the result in the form $Q(x) + \dfrac{R}{x - 2}$.

Solution. The divisor is $x - 2$, so $a = 2$. The $x^2$ term is missing, so its coefficient is $0$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrrr} 2 & 2 & -5 & 0 & 7 & -3 \\ & & 4 & -2 & -4 & 6 \\ \hline & 2 & -1 & -2 & 3 & \boxed{3} \end{array}$$

The dividend has degree $4$, so the quotient has degree $3$: $Q(x) = 2x^3 - x^2 - 2x + 3$, and $R = 3$.

$$\frac{2x^4 - 5x^3 + 7x - 3}{x - 2} = 2x^3 - x^2 - 2x + 3 + \frac{3}{x - 2}, \qquad x \ne 2$$

Example 4: Dividing by 2x − 1

Divide $4x^3 - 8x^2 + 3x + 5$ by $2x - 1$.

Solution. The leading term of the divisor is $2x$, so divide by $2x$ each time.

• $4x^3 \div 2x = 2x^2$. Multiply: $2x^2(2x - 1) = 4x^3 - 2x^2$. Subtract: $(4x^3 - 8x^2) - (4x^3 - 2x^2) = -6x^2$. Bring down $3x$.
• $-6x^2 \div 2x = -3x$. Multiply: $-3x(2x - 1) = -6x^2 + 3x$. Subtract: $(-6x^2 + 3x) - (-6x^2 + 3x) = 0$. Bring down $5$.
• $5 \div 2x$ isn’t a polynomial term, and $5$ has a lower degree than $2x - 1$, so we stop.

$$4x^3 - 8x^2 + 3x + 5 = (2x - 1)(2x^2 - 3x) + 5$$

Check: $(2x - 1)(2x^2 - 3x) = 4x^3 - 6x^2 - 2x^2 + 3x = 4x^3 - 8x^2 + 3x$, and adding $5$ gives the dividend. ✓

Aside. You can use synthetic division here with $a = \tfrac{1}{2}$ (the zero of $2x - 1$). The bottom row is $4,\ -6,\ 0 \mid 5$, which is the quotient for dividing by $x - \tfrac{1}{2}$. Since $2x - 1 = 2\left(x - \tfrac{1}{2}\right)$, you then divide those quotient coefficients by $2$ to get $2x^2 - 3x$. The remainder stays $5$.

Common mistakes

Forgetting placeholders for missing powers. In $2x^3 + 3x^2 - 4$ there’s no $x$ term, so write $0x$. Without it, the columns slide out of line and every later step is wrong. The same goes for synthetic division: write a $0$ for each missing power.

Subtracting only the first term. When you subtract $-x(x + 2) = -x^2 - 2x$, you must subtract both terms: $(-x^2 + 0x) - (-x^2 - 2x) = 2x$. Putting the subtracted line in brackets, or changing all its signs before adding, avoids this.

Using the wrong sign in synthetic division. For the divisor $x + 3$, the number in the box is $-3$, not $3$. Ask yourself: what value of $x$ makes the divisor zero?

Using synthetic division for 2x − 1 without adjusting. Synthetic division with $a = \tfrac{1}{2}$ gives the quotient for $x - \tfrac{1}{2}$, which is twice the quotient for $2x - 1$. Either divide those coefficients by $2$ or use long division.

Reading the quotient’s degree wrong. In synthetic division, the bottom row (without the remainder) starts one degree lower than the dividend. Dividing a quartic by $x - a$ gives a cubic.

Dropping the divisor from the remainder. $\dfrac{P(x)}{x - 2} = Q(x) + 3$ is wrong. The remainder must stay over the divisor: $Q(x) + \dfrac{3}{x - 2}$.

Practice

1. (Warm-up) For the statement $2x^3 - x + 5 = (x - 1)(2x^2 + 2x + 1) + 6$, name the dividend, divisor, quotient and remainder.

Solution

Dividend: $2x^3 - x + 5$. Divisor: $x - 1$. Quotient: $2x^2 + 2x + 1$. Remainder: $6$.

2. (Warm-up) Use synthetic division to divide $x^3 - 2x^2 + 4x - 5$ by $x - 2$.

Solution

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 1 & -2 & 4 & -5 \\ & & 2 & 0 & 8 \\ \hline & 1 & 0 & 4 & \boxed{3} \end{array}$$

$Q(x) = x^2 + 0x + 4 = x^2 + 4$ and $R = 3$.

3. (Warm-up) Divide $3x^2 + 5x - 1$ by $x + 1$.

Solution

$3x^2 \div x = 3x$. Subtract $3x(x + 1) = 3x^2 + 3x$ to get $2x - 1$.

$2x \div x = 2$. Subtract $2(x + 1) = 2x + 2$ to get $-3$.

$$3x^2 + 5x - 1 = (x + 1)(3x + 2) - 3$$

4. (Core) Use long division to divide $x^4 - 3x^2 + 2x - 7$ by $x + 2$.

Solution

Write the dividend as $x^4 + 0x^3 - 3x^2 + 2x - 7$.

• $x^4 \div x = x^3$. Subtract $x^3(x + 2) = x^4 + 2x^3$ to get $-2x^3$. Bring down $-3x^2$.
• $-2x^3 \div x = -2x^2$. Subtract $-2x^2(x + 2) = -2x^3 - 4x^2$ to get $x^2$. Bring down $2x$.
• $x^2 \div x = x$. Subtract $x(x + 2) = x^2 + 2x$ to get $0$. Bring down $-7$.
• $-7$ has lower degree than $x + 2$, so the remainder is $-7$.

$$x^4 - 3x^2 + 2x - 7 = (x + 2)(x^3 - 2x^2 + x) - 7$$

5. (Core) Use synthetic division to divide $x^3 + 6x^2 + 5x - 8$ by $x + 3$, and write the division statement.

Solution

The divisor is $x - (-3)$, so $a = -3$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} -3 & 1 & 6 & 5 & -8 \\ & & -3 & -9 & 12 \\ \hline & 1 & 3 & -4 & \boxed{4} \end{array}$$$$x^3 + 6x^2 + 5x - 8 = (x + 3)(x^2 + 3x - 4) + 4$$

6. (Core) Divide $6x^3 + 7x^2 - x + 3$ by $3x + 2$.

Solution

• $6x^3 \div 3x = 2x^2$. Subtract $2x^2(3x + 2) = 6x^3 + 4x^2$ to get $3x^2$. Bring down $-x$.
• $3x^2 \div 3x = x$. Subtract $x(3x + 2) = 3x^2 + 2x$ to get $-3x$. Bring down $3$.
• $-3x \div 3x = -1$. Subtract $-1(3x + 2) = -3x - 2$ to get $5$.

$$6x^3 + 7x^2 - x + 3 = (3x + 2)(2x^2 + x - 1) + 5$$

7. (Core) Write $f(x) = \dfrac{3x^3 - 2x^2 + 5}{x - 1}$ in the form $Q(x) + \dfrac{R}{x - 1}$.

Solution

Synthetic division with $a = 1$ and a $0$ for the missing $x$ term:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 3 & -2 & 0 & 5 \\ & & 3 & 1 & 1 \\ \hline & 3 & 1 & 1 & \boxed{6} \end{array}$$$$f(x) = 3x^2 + x + 1 + \frac{6}{x - 1}, \qquad x \ne 1$$

8. (Challenge) Divide $x^4 + 2x^3 - x + 3$ by $x^2 + 1$. (Hint: the remainder can be linear.)

Solution

Write the dividend as $x^4 + 2x^3 + 0x^2 - x + 3$ and the divisor as $x^2 + 0x + 1$.

• $x^4 \div x^2 = x^2$. Subtract $x^2(x^2 + 1) = x^4 + x^2$ to get $2x^3 - x^2 - x$.
• $2x^3 \div x^2 = 2x$. Subtract $2x(x^2 + 1) = 2x^3 + 2x$ to get $-x^2 - 3x + 3$.
• $-x^2 \div x^2 = -1$. Subtract $-1(x^2 + 1) = -x^2 - 1$ to get $-3x + 4$.
• $-3x + 4$ has degree $1$, lower than the divisor’s degree $2$, so we stop.

$$x^4 + 2x^3 - x + 3 = (x^2 + 1)(x^2 + 2x - 1) + (-3x + 4)$$

9. (Challenge) A box has volume $V(x) = 2x^3 + 11x^2 + 17x + 6$ cubic centimetres and height $x + 2$ centimetres. Find an expression for the area of its base, then factor it to find possible expressions for the length and width.

Solution

Volume is base area times height, so divide $V(x)$ by $x + 2$, using $a = -2$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} -2 & 2 & 11 & 17 & 6 \\ & & -4 & -14 & -6 \\ \hline & 2 & 7 & 3 & \boxed{0} \end{array}$$

The remainder is $0$, so the base area is $2x^2 + 7x + 3$ square centimetres. Factoring, $2x^2 + 7x + 3 = (2x + 1)(x + 3)$, so the base could be $2x + 1$ cm by $x + 3$ cm.