The volume of a rectangular prism is \begin{align*}2x^3 + 5x^2 - x - 6\end{align*}

### 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*}

#### Example A

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

**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 \begin{align*}2x^3-x^2-16x+15\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*}

**Remainder Theorem:** If \begin{align*}f(k) = r\end{align*}

This means that if you substitute in \begin{align*}x = k\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*}.

**Solution:** 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*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.10.