# Synthetic Division of Polynomials

The volume of a rectangular prism is 2x3+5x2−x−6. Determine if 2x+3 is the length of one of the prism's sides.

# Synthetic Division

Synthetic division is an alternative to long division. 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.

Let's use synthetic division to divide 2x4−5x3−14x2+47x−30 by x−2.

Using synthetic division, the setup is as follows:

Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is 2x3−x2−16x+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)(2x3−x2−16x+15)=2x4−5x3−14x2+47x−30.

Now, let's determine if 4 is a solution to f(x)=5x3+6x2−24x−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).

Finally, let's determine if (2x−5) is a factor of 4x4−9x2−100.

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 problem, we need zeros for the x3−term and the x−term.

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

Example 1

Earlier, you were asked to determine if 2x+3 is the length of one of the prism's sides.

Solution

If 2x+3 divides evenly into 2x3+5x2−x−6 then it is the length of one of the prism's sides.

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 2x+3=0 then x=$$\ -\frac{3}{2}$$. Therefore, we need to put $$\ -\frac{3}{2}$$ 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.

Example 2

Divide x3+9x2+12x−27 by (x+3). Write the resulting polynomial with the remainder (if there is one).

Solution

Using synthetic division, divide by -3.

The answer is $$\ x^{2}+6 x-6-\frac{9}{x+3}$$.

Example 3

Divide 2x4−11x3+12x2+9x−2 by (2x+1). Write the resulting polynomial with the remainder (if there is one).

Solution

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

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

Example 4

Is 6 a solution for f(x)=x3−8x2+72? If so, find the real-number zeros (solutions) of the resulting polynomial.

Solution

Put a zero placeholder for the x−term. Divide by 6. The resulting polynomial is x2−2x−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 $$\ 6,1+\sqrt{13} \approx 4.61 \text { and } 1-\sqrt{13} \approx-2.61$$.

# Review

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

1. (x3+6x2+7x+10)÷(x+2)
2. (4x3−15x2−120x−128)÷(x−8)
3. (4x2−5)÷(2x+1)
4. (2x4−15x3−30x2−20x+42)÷(x+9)
5. (x3−3x2−11x+5)÷(x−5)
6. (3x5+4x3−x−2)÷(x−1)
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 f(−2) if f(x)=2x4−5x3−10x2+21x−4.
10. Now, divide 2x4−5x3−10x2+21x−4 by (x+2) synthetically. What do you notice?

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

1. 12x3+76x2+107x−20; −4
2. x3−5x2−2x+10; −2
3. 6x3−17x2+11x−2; 2

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

1. x4+7x3+6x2−32x−32; −4, −1
2. 6x4+19x3+11x2−6x; 0, −2