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Synthetic Division of Polynomials

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Synthetic Division of Polynomials

If you have completed the lesson on oblique asymptotes, you probably know that the problem:

Find the oblique asymptote of f(x)=\frac{3x^{3}-2x^{2}+11x-9}{x^{2}+2}

could be an excellent reason to groan and grumble due to the long division that would be required.

Isn't there a better way?

Watch This

James Sousa: Polynomial Division: Synthetic Division


Synthetic division is an alternative to long division from the previous concept. It can also be used to divide a polynomial by a possible factor, x - k . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

Example A

Divide 2x^4-5x^3-14x^2+47x-30 by x - 2 .

Solution: Using synthetic division, the setup is as follows:

To “read” the answer, use the numbers as follows:

Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is 2x^3-x^2-16x+15 . Notice that when we synthetically divide by k , the “leftover” polynomial is one degree less than the original. We could also write (x-2)(2x^3-x^2-16x+15)=2x^4-5x^3-14x^2+47x-30 .

Example B

Determine if 4 is a solution to f(x)=5x^3+6x^2-24x-16 .

Using synthetic division, we have:

The remainder is 304, so 4 is not a solution. Notice if we substitute in x = 4 , also written f(4) , we would have f(4)=5(4)^3+6(4)^2-24(4)-16=304 . This leads us to the Remainder Theorem.

Remainder Theorem: If f(k) = r , then r is also the remainder when dividing by (x - k) .

This means that if you substitute in x = k or divide by k , what comes out of f(x) is the same. r is the remainder, but it is also the corresponding y- value. Therefore, the point (k, r) would be on the graph of f(x) .

Example C

Determine if (2x - 5) is a factor of 4x^4-9x^2-100 .

Solution: If you use synthetic division, the factor is not in the form (x - k) . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put \frac{5}{2} up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this example, we need zeros for the x^3- term and the x- term.

This means that \frac{5}{2} is a zero and its corresponding binomial, (2x - 5) , is a factor.

Intro Problem Revisit

If we want to use synthetic division, notice that the factor is not in the form (x - k) . Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If x^2 + 2 = 0 , then x = \pm i \sqrt{2} , which can also be expressed as x = +i \sqrt{2} or x = -i \sqrt{2} . Therefore, we need to use synthetic division twice because there are two complex roots .

To start, put i \sqrt{2} up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of -5 + 5 i \sqrt{2} .

Next, we divide results from the last synthetic division with the other complex root. Put -i \sqrt{2} up in the left-hand corner box.

As a result, f(x) = \frac{3x^3 - 2x^2 + 11x - 9}{x^2 + 2} = 3x - 2 + \frac{5x - 5}{x^2 + 2} .

Guided Practice

1. Divide x^3+9x^2+12x-27 by (x + 3) . Write the resulting polynomial with the remainder (if there is one).

2. Is 6 a solution for f(x)=x^3-8x^2+72 ? If so, find the real-number zeros (solutions) of the resulting polynomial.

3. Divide 4x^3+3x^2+10x+4 by (x^2 + 2) . Write the resulting polynomial with the remainder (if there is one).


1. Using synthetic division, divide by -\frac{1}{2} .

The answer is 2x^3-12x^2+18x-\frac{2}{2x+1} .

2. Put a zero placeholder for the x- term. Divide by 6.

The resulting polynomial is x^2-2x-12 . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

x=\frac{2 \pm \sqrt{2^2-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}

The solutions to this polynomial are 6, 1+\sqrt{13} \approx 4.61 and 1-\sqrt{13} \approx -2.61 .

3. Using synthetic division, divide by i \sqrt{2} .

Divide the results from the last step by -i \sqrt{2} .

The answer is 4x + 3 + \frac{2x - 2}{x^2 + 2} .


Synthetic Division
An alternative to long division for dividing f(x) by k where only the coefficients of f(x) are used.
Remainder Theorem
If f(k) = r , then r is also the remainder when dividing by (x - k) .

Explore More

Divide using synthetic division:

  1. \frac{7x^2 - 23x + 6}{x - 3}
  2. \frac{x^4 - 5x + 10}{x + 3}
  3. (2x^2 + 13x - 8) \div (x - \frac{1}{2})
  4. (x^4 + 6x^3 + 6x^2) \div (x + 5)
  5. \frac{x^3 - 7x - 6}{x + 2}
  6. \frac{8y^3 + y^4 + 16 + 32y + 24y^2}{y + 2}

Use synthetic substitution to evaluate the polynomial function for the given value:

  1. P(x) = 2x^2 - 5x - 3 for x = 4
  2. P(x) = 4x^3 - 5x^2 + 3 for x = -1
  3. p(x) = 3x^3 - 5x^2 - x =2 for x = -\frac{1}{3}
  4. The area of a rectangle is  3x^3 - 11x^2 - 56x - 48 and the length is 3x + 4 . What is the width?
  5. A group of geologists have taken a collection of samples of a substance from a proposed mining site and must identify the substance. Each sample is roughly cylindrical. The volume of each sample as a function of cylinder height (in centimeters) is V(h) = \frac {1}{4} \pi h^3 . The mass (in grams) of each sample in terms of height can be modeled by M(h) = \frac{1}{4}h^3 - h^2 + 5h . Write an expression that represents the density of the samples. (Hint: D = \frac{M}{V} )

Divide using synthetic division:

  1. 4x^3 - 3x^2 + x + 1 divided by (x^2 - 1)
  2. x^4 - x^2 + 1 divided by (x^2 + 1) )
  3. 2x^4 - \frac{1}{2}x^2 + 2x divided by (4x^2 - 1)
  4. 4x^3 - 3x^2 + x + 1 divided by (4x^2 + 1)

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