<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

6.10: Synthetic Division of Polynomials

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated21 minsto complete
%
Progress
Practice Synthetic Division of Polynomials
 
 
 
MEMORY METER
This indicates how strong in your memory this concept is
Practice
Progress
Estimated21 minsto complete
%
Estimated21 minsto complete
%
Practice Now
MEMORY METER
This indicates how strong in your memory this concept is
Turn In

The volume of a rectangular prism is 2x3+5x2x6. Determine if 2x+3 is the length of one of the prism's sides.

Watch This

James Sousa: Polynomial Division: Synthetic Division

Guidance

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, xk. However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

Example A

Divide 2x45x314x237x30 by x2.

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 2x3x216x+15. Notice that when we synthetically divide by k, the “leftover” polynomial is one degree less than the original. We could also write (x2)(2x3x216x+15)=2x45x314x2+47x30.

Example B

Determine if 4 is a solution to f(x)=5x3+6x224x16.

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)224(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 (xk).

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 also is the corresponding yvalue. Therefore, the point (k,r) would be on the graph of f(x).

Example C

Determine if (2x5) is a factor of 4x49x2100.

Solution: If you use synthetic division, the factor is not in the form (xk). We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put 52 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 x3term and the xterm.

This means that 52 is a zero and its corresponding binomial, (2x5), is a factor.

Intro Problem Revisit

If 2x+3 divides evenly into 2x3+5x2x6 then it is the length of one of the prism's sides.

If we use synthetic division, the factor is not in the form (xk). We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put 32 up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of 0. This means that (2x+3) is a factor of the volume. Therefore, it is also the length of one of the sides of the rectangular prism.

Guided Practice

1. Divide x39x2+12x27 by (x+3). Write the resulting polynomial with the remainder (if there is one).

2. Divide 2x411x3+12x2+9x2 by (2x+1). Write the resulting polynomial with the remainder (if there is one).

3. Is 6 a solution for f(x)=x38x2+72? If so, find the real-number zeros (solutions) of the resulting polynomial.

Answers

1. Using synthetic division, divide by -3.

The answer is x2+6x69x+3.

2. Using synthetic division, divide by 12.

The answer is 2x312x2+18x22x+1.

3. Put a zero placeholder for the xterm. Divide by 6.

The resulting polynomial is x22x12. While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

x=2±224(1)(12)2=2±4+482=2±2132=1±13

The solutions to this polynomial are 6, 1+134.61 and 1132.61.

Vocabulary

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

Practice

Use synthetic division to divide the following polynomials. Write out the remaining polynomial.

  1. \begin{align*}(x^3+6x^2+7x+10) \div (x+2)\end{align*}
  2. \begin{align*}(4x^3-15x^2-120x-128) \div (x-8)\end{align*}
  3. \begin{align*}(4x^2-5) \div (2x+1)\end{align*}
  4. \begin{align*}(2x^4-15x^3-30x^2-20x+42) \div (x+9)\end{align*}
  5. \begin{align*}(x^3-3x^2-11x+5) \div (x-5)\end{align*}
  6. \begin{align*}(3x^5+4x^3-x-2) \div (x-1)\end{align*}
  7. Which of the division problems above generate no remainder? What does that mean?
  8. What is the difference between a zero and a factor?
  9. Find \begin{align*}f(-2)\end{align*} if \begin{align*}f(x)=2x^4-5x^3-10x^2+21x-4\end{align*}.
  10. Now, divide \begin{align*}2x^4-5x^3-10x^2+21x-4\end{align*} by \begin{align*}(x + 2)\end{align*} synthetically. What do you notice?

Find all real zeros of the following polynomials, given one zero.

  1. \begin{align*}12x^3+76x^2+107x-20; -4\end{align*}
  2. \begin{align*}x^3-5x^2-2x+10; -2\end{align*}
  3. \begin{align*}6x^3-17x^2+11x-2; 2\end{align*}

Find all real zeros of the following polynomials, given two zeros.

  1. \begin{align*}x^4+7x^3+6x^2-32x-32; -4, 1\end{align*}
  2. \begin{align*}6x^4+19x^3+11x-6; 0, -2\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Vocabulary

TermDefinition
Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Remainder Theorem The remainder theorem states that if f(k) = r, then r is the remainder when dividing f(x) by (x - k).
Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.

Image Attributions

Show Hide Details
Description
Difficulty Level:
At Grade
Grades:
Date Created:
Mar 12, 2013
Last Modified:
Sep 07, 2016
Files can only be attached to the latest version of Modality
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.ALY.252.L.2