Synthetic Division Examples (page 2 of 4)

Sections: Introduction, Worked examples, Finding zeroes, Factoring polynomials

• Complete the indicated division.

For this first exercise, I will display the entire synthetic-division process step-by-step.

First, carry down the "2" that indicates the leading coefficient:
Multiply by the number on the left, and carry the result into the next column:
Add down the column:
Multiply by the number on the left, and carry the result into the next column:
Add down the column:
Multiply by the number on the left, and carry the result into the next column:
Add down the column:
Multiply by the number on the left, and carry the result into the next column:
Add down the column for the remainder:
The completed division is:

This exercise never said anything about polynomials, factors, or zeroes, but this division says that, if you divide 2x⁴ – 3x³ – 5x² + 3x + 8 by x – 2, then the remainder will be 2, and therefore x – 2 is not a factor of 2x⁴ – 3x³ – 5x² + 3x + 8, and x = 2 is not a zero (that is, a root or x-intercept) of the initial polynomial. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Divide 3x³ – 2x² + 3x – 4 by x – 3 using synthetic division.
  Write the answer in the form " q(x) + ^r(x)/_d(x) ".

This question is asking me, in effect, to convert an "improper" polynomial "fraction" into a polynomial "mixed number". That is, I am being asked to do something similar to converting the improper fraction ¹⁷/₅ to the mixed number 3 ²/₅, which is really the shorthand for the addition expression "3 + ²/₅".

To convert the polynomial division into the required "mixed number" format, I have to do the division; I will show most of the steps.

First, write down all the coefficients, and put the zero from x – 3 = 0 (so x = 3) at the left.
Next, carry down the leading coefficient:
Multiply by the potential zero, carry up to the next column, and add down:
Repeat this process:
Repeat this process again:

As you can see, the remainder is 68. Since I started with a polynomial of degree 3 and then divided by x – 3 (that is, by a polynomial of degree 1), I am left with a polynomial of degree 2. Then the bottom line represents the polynomial 3x² + 7x + 24 with a remainder of 68. Putting this result into the required "mixed number" format, I get the answer as being:

It is always true that, when you use synthetic division, your answer (in the bottom row) will be of degree one less than what you'd started with, because you have divided out a linear factor. That was how I knew that my answer, denoted by the "3  7  24" in the bottom row, stood for a quadratic, since I had started with a cubic.

<< Previous  Top  |  1 | 2 | 3 | 4  |  Return to Index  Next >>

  Copyright © 2002-2012  Elizabeth Stapel   |   About   |   Terms of Use

Contact Us