Polynomial long division and synthetic division

Q: When can I use synthetic division?

Only when you divide by a linear factor of the form x - c, like x - 4 or x + 3. For a divisor such as x^2 + 2x - 1, use long division. For 2x - 1, use c = \frac{1}{2} and then divide the quotient by 2.

Q: What is the difference between the remainder theorem and the factor theorem?

The remainder theorem says dividing p(x) by x - c leaves a remainder of p(c). The factor theorem is the special case where that remainder is 0: then x - c divides evenly and is a factor, and c is a root.

Q: How do I write the answer when there is a remainder?

Write the quotient plus the remainder over the divisor: x^2 - x - 1 + \frac{2}{x - 3}. You can also write p(x) = (x - 3)(x^2 - x - 1) + 2. Both say the same thing. Your teacher may prefer one form, so check how your class writes it.

The key idea

Dividing polynomials works like long division with numbers. You get a quotient and a remainder, and they always fit this equation:

p(x) = d(x)\,q(x) + r(x)

Here p is what you divide, d is the divisor, q is the quotient and r is the remainder, which has a lower degree than d. When you divide by $x - c$ , the remainder is a single number, and plugging in $x = c$ shows it equals $p(c)$ . That is the remainder theorem.

Worked examples

Example 1: long division with a remainder

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

Divide the leading terms: $x^3 \div x = x^2$ . Multiply $x^2$ by the divisor.
$x^2(x - 3) = x^3 - 3x^2$
Subtract that from the first two terms.
$(x^3 - 4x^2) - (x^3 - 3x^2) = -x^2$
Bring down $2x$ to get $-x^2 + 2x$ . Divide: $-x^2 \div x = -x$ . Multiply.
$-x(x - 3) = -x^2 + 3x$
Subtract.
$(-x^2 + 2x) - (-x^2 + 3x) = -x$
Bring down 5 to get $-x + 5$ . Divide: $-x \div x = -1$ . Multiply and subtract.
$(-x + 5) - (-1)(x - 3) = 2$
Nothing is left to bring down, so 2 is the remainder. Check by multiplying back.
$(x - 3)(x^2 - x - 1) + 2 = x^3 - 4x^2 + 2x + 5$

Answer Quotient $x^2 - x - 1$ , remainder 2. So $\frac{x^3 - 4x^2 + 2x + 5}{x - 3} = x^2 - x - 1 + \frac{2}{x - 3}$ .

Example 2: synthetic division, then factor

Problem Divide $2x^3 + 3x^2 - 11x - 6$ by $x - 2$ and factor the polynomial completely.

For $x - 2$ , use $c = 2$ . Write the coefficients 2, 3, $-11$ , $-6$ . Bring down the first one: 2.
Multiply by c and add to the next coefficient.
$3 + 2 \cdot 2 = 7$
Repeat with the new number.
$-11 + 7 \cdot 2 = 3$
Repeat once more. The last number is the remainder.
$-6 + 3 \cdot 2 = 0$
The bottom row 2, 7, 3 gives the quotient $2x^2 + 7x + 3$ (one degree lower). The remainder is 0, so $x - 2$ is a factor. Factor the quotient.
$2x^2 + 7x + 3 = (2x + 1)(x + 3)$

Answer Quotient $2x^2 + 7x + 3$ , remainder 0, so $2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x + 1)(x + 3)$ .

The synthetic division layout for Example 2 (c = 2)
	$x^3$	$x^2$	$x$	constant
Coefficients	2	3	$-11$	$-6$
Add c times the number before		4	14	6
Bottom row	2	7	3	0 (remainder)

Example 3: the remainder theorem

Problem Find the remainder when $p(x) = x^4 - 3x^3 + 5x - 8$ is divided by $x + 2$ .

$x + 2$ is $x - (-2)$ , so $c = -2$ . By the remainder theorem, the remainder is $p(-2)$ . No division needed.
Evaluate carefully with parentheses.
$(-2)^4 - 3(-2)^3 + 5(-2) - 8 = 16 + 24 - 10 - 8 = 22$

Answer The remainder is 22.

Example 4 (test-hard): find k so that x - 3 is a factor

Problem For what value of k is $x - 3$ a factor of $p(x) = x^3 + kx^2 - 4x + 12$ ? Then factor $p(x)$ completely.

By the factor theorem, $p(3)$ must be 0.
$27 + 9k - 12 + 12 = 0$
Simplify.
$27 + 9k = 0$
Solve.
$k = -3$
Now $p(x) = x^3 - 3x^2 - 4x + 12$ . Factor by grouping.
$x^3 - 3x^2 - 4x + 12 = x^2(x - 3) - 4(x - 3)$
Pull out $x - 3$ and use the difference of squares.
$x^2(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2)$

Answer $k = -3$ , and $p(x) = (x - 3)(x - 2)(x + 2)$ .

Common mistakes

Skipping a missing power. Dividing $3x^3 - 5x^2 + 4$ needs coefficients 3, $-5$ , 0, 4. Fix: write a 0 for every missing power before you start.
Using the wrong sign for c. For $x + 2$ , c is $-2$ , not 2. Fix: set the divisor equal to zero and solve: $x + 2 = 0$ gives $x = -2$ .
Subtracting only the first term in long division. $(-x^2 + 2x) - (-x^2 + 3x)$ is $-x$ , not $5x$ . Fix: put the whole product in parentheses and distribute the minus sign.
Reading the quotient's degree wrong. Dividing a cubic by $x - c$ gives a quadratic. Fix: the bottom row starts one power lower than you began with.
Using synthetic division for a divisor like $x^2 + 1$ . Synthetic division only works for divisors of the form $x - c$ . Fix: use long division for anything else.

Quick methods

Practice

5 practice questions

What is the remainder when $x^3 + 2x^2 - 5x + 1$ is divided by $x - 1$ ?
1. $-1$
2. $1$
3. $7$
4. $9$
Show answer
Answer: $-1$
By the remainder theorem, compute $p(1) = 1 + 2 - 5 + 1 = -1$ . The answer 7 is $p(-1)$ , which uses the wrong sign for c.
Divide $3x^3 - 5x^2 + 4$ by $x - 2$ . What are the quotient and remainder?
1. $3x^2 + x + 2$ , remainder $8$
2. $3x^2 - 11x + 22$ , remainder $-40$
3. $3x + 1$ , remainder $6$
4. $3x^2 + x + 2$ , remainder $0$
Show answer
Answer: $3x^2 + x + 2$ , remainder $8$
Coefficients 3, $-5$ , 0, 4 with $c = 2$ give the bottom row 3, 1, 2, 8. The remainder 8 also equals $p(2) = 24 - 20 + 4$ . The $-40$ choice used $c = -2$ , and $3x + 1$ skipped the 0 placeholder.
Which of these is a factor of $p(x) = x^3 - 7x - 6$ ?
1. $x - 1$
2. $x - 2$
3. $x + 2$
4. $x + 3$
Show answer
Answer: $x + 2$
$p(-2) = -8 + 14 - 6 = 0$ , so $x + 2$ is a factor. The others give $p(1) = -12$ , $p(2) = -12$ and $p(-3) = -12$ , so they are not factors.
When $p(x) = x^3 - 2x^2 + kx + 6$ is divided by $x - 2$ , the remainder is 4. What is k?
1. $-5$
2. $-1$
3. $1$
4. $5$
Show answer
Answer: $-1$
$p(2) = 8 - 8 + 2k + 6 = 2k + 6$ . Set it equal to 4: $2k = -2$ , so $k = -1$ . Setting $p(2) = 0$ instead would give $k = -3$ , which is not the question.
What is the remainder when $2x^4 - x^3 + 3$ is divided by $x + 1$ ?
Show answer
Answer: 6
Use $c = -1$ : $2(-1)^4 - (-1)^3 + 3 = 2 + 1 + 3 = 6$ . Watch the odd power: $(-1)^3 = -1$ , and subtracting it adds 1.

Frequently asked questions

When can I use synthetic division?

Only when you divide by a linear factor of the form $x - c$ , like $x - 4$ or $x + 3$ . For a divisor such as $x^2 + 2x - 1$ , use long division. For $2x - 1$ , use $c = \frac{1}{2}$ and then divide the quotient by 2.

What is the difference between the remainder theorem and the factor theorem?

The remainder theorem says dividing $p(x)$ by $x - c$ leaves a remainder of $p(c)$ . The factor theorem is the special case where that remainder is 0: then $x - c$ divides evenly and is a factor, and c is a root.

How do I write the answer when there is a remainder?

Write the quotient plus the remainder over the divisor: $x^2 - x - 1 + \frac{2}{x - 3}$ . You can also write $p(x) = (x - 3)(x^2 - x - 1) + 2$ . Both say the same thing. Your teacher may prefer one form, so check how your class writes it.

How do I find a root to start factoring a cubic?

Try the rational root test: possible rational roots are factors of the constant term divided by factors of the leading coefficient. Test them with $p(c)$ or synthetic division until one gives 0, then factor the quadratic that is left.

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

Related

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