Grade 12 · MHF4U · Advanced Functions

MHF4U: Long and Synthetic Division

Dividing polynomials is the same move as long division with numbers: take out as many copies of the divisor as you can, and whatever is left is the remainder. Synthetic division is the shortcut for a linear divisor.

How It Works

Every division ends in a division statement: dividend = divisor × quotient + remainder, or $P(x) = D(x)Q(x) + R(x)$ . Write it at the end of every question. It is the answer, and expanding it is how you check your work.

Long division works for any divisor. Write the dividend in descending powers and put in a $0$ for every missing power, so $x^3 - 7x + 2$ becomes $x^3 + 0x^2 - 7x + 2$ . Then repeat four steps: divide the leading terms, multiply that term by the divisor, subtract, bring down the next term. Stop when the remainder's degree is lower than the divisor's.

Synthetic division is faster, but only for a linear divisor. For $x - k$ , use $k$ : so $x - 2$ means $2$ and $x + 3$ means $-3$ . List the coefficients, zeros included, and bring the first one down. Multiply it by $k$ , add the product to the next coefficient, and keep going. The last number is the remainder; the others are the quotient's coefficients, one degree lower than the dividend.

Synthetic division does not warn you about a missing term, so the zeros are on you. And when the divisor is $ax - b$ with $a \ne 1$ , use $\frac{b}{a}$ , then divide every quotient coefficient by $a$ . The remainder stays as it is.

Quick check on any answer: the quotient's degree plus the divisor's degree equals the dividend's degree.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Divide $x^2 + 5x + 7$ by $x + 2$ using synthetic division.
Show solution
1. The divisor is $x + 2$ , so use $k = -2$ . The coefficients are $1, 5, 7$ .
2. Bring down the $1$ . Multiply: $1 \times (-2) = -2$ . Add to $5$ : $3$ .
3. Multiply: $3 \times (-2) = -6$ . Add to $7$ : $1$ .
4. The bottom row is $1, 3, 1$ . The last number is the remainder, so the quotient is $x + 3$ and the remainder is $1$ .
Answer: $x^2 + 5x + 7 = (x + 2)(x + 3) + 1$
Easy
Divide $2x^3 - 3x^2 - 4x + 5$ by $x - 2$ using synthetic division.
Show solution
1. Use $k = 2$ . The coefficients are $2, -3, -4, 5$ .
2. Bring down $2$ . Then $2 \times 2 = 4$ and $-3 + 4 = 1$ .
3. $1 \times 2 = 2$ and $-4 + 2 = -2$ .
4. $-2 \times 2 = -4$ and $5 + (-4) = 1$ .
5. The bottom row $2, 1, -2, 1$ gives the quotient $2x^2 + x - 2$ and remainder $1$ .
Answer: $2x^3 - 3x^2 - 4x + 5 = (x - 2)(2x^2 + x - 2) + 1$
Medium
Divide $x^4 - 3x^2 + 2x - 6$ by $x + 3$ .
Show solution
1. There is no $x^3$ term, so the coefficients are $1, 0, -3, 2, -6$ . Use $k = -3$ .
2. Bring down $1$ . Then $1 \times (-3) = -3$ and $0 + (-3) = -3$ .
3. $-3 \times (-3) = 9$ and $-3 + 9 = 6$ .
4. $6 \times (-3) = -18$ and $2 + (-18) = -16$ .
5. $-16 \times (-3) = 48$ and $-6 + 48 = 42$ .
6. The dividend has degree 4, so the quotient has degree 3: $x^3 - 3x^2 + 6x - 16$ , remainder $42$ .
Answer: $x^4 - 3x^2 + 2x - 6 = (x + 3)(x^3 - 3x^2 + 6x - 16) + 42$
Medium
Divide $2x^3 + x^2 - 5x + 4$ by $x^2 + x - 1$ .
Show solution
1. The divisor is quadratic, so this is long division.
2. Divide the leading terms: $\frac{2x^3}{x^2} = 2x$ . Multiply: $2x(x^2 + x - 1) = 2x^3 + 2x^2 - 2x$ .
3. Subtract: $(2x^3 + x^2 - 5x + 4) - (2x^3 + 2x^2 - 2x) = -x^2 - 3x + 4$ .
4. Divide again: $\frac{-x^2}{x^2} = -1$ . Multiply: $-1(x^2 + x - 1) = -x^2 - x + 1$ .
5. Subtract: $(-x^2 - 3x + 4) - (-x^2 - x + 1) = -2x + 3$ .
6. $-2x + 3$ has degree 1, lower than the divisor's degree 2, so stop. Quotient $2x - 1$ , remainder $-2x + 3$ .
Answer: $2x^3 + x^2 - 5x + 4 = (x^2 + x - 1)(2x - 1) + (-2x + 3)$
Test level
Divide $6x^3 + 5x^2 - 8x + 3$ by $2x - 1$ using synthetic division, and write the division statement.
Show solution
1. $2x - 1 = 0$ at $x = \frac{1}{2}$ , so use $k = \frac{1}{2}$ . The coefficients are $6, 5, -8, 3$ .
2. Bring down $6$ . Then $6 \times \frac{1}{2} = 3$ and $5 + 3 = 8$ .
3. $8 \times \frac{1}{2} = 4$ and $-8 + 4 = -4$ .
4. $-4 \times \frac{1}{2} = -2$ and $3 + (-2) = 1$ .
5. That row divides by $x - \frac{1}{2}$ : $6x^3 + 5x^2 - 8x + 3 = (x - \frac{1}{2})(6x^2 + 8x - 4) + 1$ .
6. The divisor is $2(x - \frac{1}{2})$ , so divide each quotient coefficient by $2$ : $3x^2 + 4x - 2$ . The remainder stays $1$ .
Answer: $6x^3 + 5x^2 - 8x + 3 = (2x - 1)(3x^2 + 4x - 2) + 1$

Learn It on Martian Lab

Martian Lab teaches this in the Polynomial Operations lesson of its MHF4U Polynomial Equations unit. Watch it, then take the Polynomial Equations practice tests to see if it stuck.

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Free Polynomial Equations practice test, with solutions

More MHF4U topics: Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations

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