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# Synthetic Division of Polynomials

## Concise method of dividing polynomials when the divisor is x minus a constant.

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Synthetic Division of Polynomials

If you have completed the lesson on oblique asymptotes, you probably know that the problem:

Find the oblique asymptote of \begin{align*}f(x)=\frac{3x^{3}-2x^{2}+11x-9}{x^{2}+2}\end{align*}

could be an excellent reason to groan and grumble due to the long division that would be required.

Isn't there a better way?

### Synthetic Division of Polynomials

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.

### Examples

#### Example 1

Earlier, you were given a question about finding a better way to divide polynomials.

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*}x^2 + 2 = 0\end{align*}, then \begin{align*}x = \pm i \sqrt{2}\end{align*}, which can also be expressed as \begin{align*}x = +i \sqrt{2}\end{align*} or \begin{align*}x = -i \sqrt{2}\end{align*}. Therefore, we need to use synthetic division twice because there are two complex roots.

To start, put \begin{align*}i \sqrt{2}\end{align*} up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of \begin{align*}-5 + 5 i \sqrt{2}\end{align*}.

Next, we divide results from the last synthetic division with the other complex root. Put \begin{align*}-i \sqrt{2}\end{align*} up in the left-hand corner box.

As a result, \begin{align*}f(x) = \frac{3x^3 - 2x^2 + 11x - 9}{x^2 + 2} = 3x - 2 + \frac{5x - 5}{x^2 + 2}\end{align*}.

#### Example 2

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

Using synthetic division, the setup is 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 3

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.

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*}.

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 4

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

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.

#### Example 5

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.

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*}.

#### Example 6

Divide \begin{align*}4x^3+3x^2+10x+4\end{align*} by \begin{align*}(x^2 + 2)\end{align*}. Write the resulting polynomial with the remainder (if there is one).

Using synthetic division, divide by \begin{align*}i \sqrt{2}\end{align*}.

Divide the results from the last step by \begin{align*}-i \sqrt{2}\end{align*}.

The answer is \begin{align*}4x + 3 + \frac{2x - 2}{x^2 + 2}\end{align*}.

### Review

Divide using synthetic division:

1. \begin{align*}\frac{7x^2 - 23x + 6}{x - 3}\end{align*}
2. \begin{align*}\frac{x^4 - 5x + 10}{x + 3}\end{align*}
3. \begin{align*}(2x^2 + 13x - 8) \div (x - \frac{1}{2})\end{align*}
4. \begin{align*}(x^4 + 6x^3 + 6x^2) \div (x + 5)\end{align*}
5. \begin{align*}\frac{x^3 - 7x - 6}{x + 2}\end{align*}
6. \begin{align*}\frac{8y^3 + y^4 + 16 + 32y + 24y^2}{y + 2}\end{align*}

Use synthetic substitution to evaluate the polynomial function for the given value:

1. \begin{align*}P(x) = 2x^2 - 5x - 3\end{align*} for \begin{align*}x = 4\end{align*}
2. \begin{align*}P(x) = 4x^3 - 5x^2 + 3 \end{align*} for \begin{align*}x = -1\end{align*}
3. \begin{align*}p(x) = 3x^3 - 5x^2 - x =2\end{align*} for \begin{align*}x = -\frac{1}{3}\end{align*}
4. The area of a rectangle is \begin{align*} 3x^3 - 11x^2 - 56x - 48\end{align*} and the length is \begin{align*}3x + 4\end{align*}. What is the width?
5. A group of geologists have taken a collection of samples of a substance from a proposed mining site and must identify the substance. Each sample is roughly cylindrical. The volume of each sample as a function of cylinder height (in centimeters) is \begin{align*}V(h) = \frac {1}{4} \pi h^3\end{align*}. The mass (in grams) of each sample in terms of height can be modeled by \begin{align*}M(h) = \frac{1}{4}h^3 - h^2 + 5h\end{align*}. Write an expression that represents the density of the samples. (Hint: \begin{align*}D = \frac{M}{V}\end{align*})

Divide using synthetic division:

1. \begin{align*}4x^3 - 3x^2 + x + 1\end{align*} divided by \begin{align*}(x^2 - 1)\end{align*}
2. \begin{align*}x^4 - x^2 + 1\end{align*} divided by \begin{align*}(x^2 + 1)\end{align*}
3. \begin{align*}2x^4 - \frac{1}{2}x^2 + 2x\end{align*} divided by \begin{align*}(4x^2 - 1)\end{align*}
4. \begin{align*}4x^3 - 3x^2 + x + 1\end{align*} divided by \begin{align*}(4x^2 + 1)\end{align*}

To see the Review answers, open this PDF file and look for section 2.11.

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### Vocabulary Language: English

Dividend

In a division problem, the dividend is the number or expression that is being divided.

divisor

In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.

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.

Polynomial long division

Polynomial long division is the standard method of long division, applied to the division of polynomials.

Quotient

The quotient is the result after two amounts have been divided.

Remainder

A remainder is the value left over if the divisor does not divide evenly into the dividend.

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