The volume of a rectangular prism is \begin{align*}2x^3 + 5x^2 - x - 6\end{align*} . Determine if @$\begin{align*}2x + 3\end{align*}@$ 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, @$\begin{align*}x - k\end{align*}@$ . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

#### Example A

Divide @$\begin{align*}2x^4-5x^3-14x^2+47x-30\end{align*}@$ by @$\begin{align*}x - 2\end{align*}@$ .

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Solution:
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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 @$\begin{align*}2x^3-x^2-16x+15\end{align*}@$ . Notice that when we synthetically divide by @$\begin{align*}k\end{align*}@$ , the “leftover” polynomial is one degree less than the original. We could also write @$\begin{align*}(x-2)(2x^3-x^2-16x+15)=2x^4-5x^3-14x^2+47x-30\end{align*}@$ .

#### Example B

Determine if 4 is a solution to @$\begin{align*}f(x)=5x^3+6x^2-24x-16\end{align*}@$ .

Using synthetic division, we have:

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

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Remainder Theorem:
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If
@$\begin{align*}f(k) = r\end{align*}@$
, then
@$\begin{align*}r\end{align*}@$
is also the remainder when dividing by
@$\begin{align*}(x - k)\end{align*}@$
.

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

#### Example C

Determine if @$\begin{align*}(2x - 5)\end{align*}@$ is a factor of @$\begin{align*}4x^4-9x^2-100\end{align*}@$ .

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Solution:
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If you use synthetic division, the factor is not in the form
@$\begin{align*}(x - k)\end{align*}@$
. We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put
@$\begin{align*}\frac{5}{2}\end{align*}@$
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
@$\begin{align*}x^3-\end{align*}@$
term and the
@$\begin{align*}x-\end{align*}@$
term.

This means that @$\begin{align*}\frac{5}{2}\end{align*}@$ is a zero and its corresponding binomial, @$\begin{align*}(2x - 5)\end{align*}@$ , is a factor.

#### Intro Problem Revisit

If @$\begin{align*}2x + 3\end{align*}@$ divides evenly into @$\begin{align*}2x^3 + 5x^2 - x - 6\end{align*}@$ 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 @$\begin{align*}(x - k)\end{align*}@$ . Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If @$\begin{align*}2x+3=0\end{align*}@$ then @$\begin{align*}x=-\frac{3}{2}\end{align*}@$ . Therefore, we need to put @$\begin{align*}\frac{-3}{2}\end{align*}@$ up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of 0. This means that @$\begin{align*}(2x + 3)\end{align*}@$ 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 @$\begin{align*}x^3+9x^2+12x-27\end{align*}@$ by @$\begin{align*}(x + 3)\end{align*}@$ . Write the resulting polynomial with the remainder (if there is one).

2. Divide @$\begin{align*}2x^4-11x^3+12x^2+9x-2\end{align*}@$ by @$\begin{align*}(2x + 1)\end{align*}@$ . Write the resulting polynomial with the remainder (if there is one).

3. Is 6 a solution for @$\begin{align*}f(x)=x^3-8x^2+72\end{align*}@$ ? If so, find the real-number zeros (solutions) of the resulting polynomial.

#### Answers

1. Using synthetic division, divide by -3.

The answer is @$\begin{align*}x^2+6x-6-\frac{9}{x+3}\end{align*}@$ .

2. Using synthetic division, divide by @$\begin{align*}-\frac{1}{2}\end{align*}@$ .

The answer is @$\begin{align*}2x^3-12x^2+18x-\frac{2}{2x+1}\end{align*}@$ .

3. Put a zero placeholder for the @$\begin{align*}x-\end{align*}@$ term. Divide by 6.

The resulting polynomial is @$\begin{align*}x^2-2x-12\end{align*}@$ . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

@$$\begin{align*}x=\frac{2 \pm \sqrt{2^2-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}\end{align*}@$$

The solutions to this polynomial are 6, @$\begin{align*}1+\sqrt{13} \approx 4.61\end{align*}@$ and @$\begin{align*}1-\sqrt{13} \approx -2.61\end{align*}@$ .

### Explore More

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

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

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

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

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