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2.11: Synthetic Division of Polynomials

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

Find the oblique asymptote of f(x)=\frac{3x^{3}-5x^{2}+2}{x^{2}-3x}

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

Isn't there a better way?

Watch This

Embedded Video:

- James Sousa: Polynomial Division: Synthetic Division

Guidance

Synthetic Division

In this lesson, we explore synthetic division , a derivative of polynomial long division .

To illustrate the value of synthetic division, we will first solve an example with long division and then redo the division by the synthetic division method.

Example A

Divide: f(x)=x^{3}+x^{2}-10x+13 by D(x)=x-2 .

Solution A: long division

& \qquad \qquad \qquad \quad \ \ x^2 + 3x - 4 \qquad \qquad \quad \leftarrow\text{Quotient}\\& \text{Divisor}\rightarrow x-2 \ \big ) \overline{x^{3} +x^{2} -10x +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \qquad \qquad \quad \ \ x^{3} -2x^{2}\\& \qquad \qquad \qquad \quad \ \ \searrow\\& \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-10x\\& \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-6x\\& \qquad \qquad \qquad \qquad \qquad \searrow\\& \qquad \qquad \qquad \qquad \qquad \quad \ -4x+13\\& \qquad \qquad \qquad \qquad \qquad \quad \ \ -4x+8\\& \qquad \qquad \qquad \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad +5 \qquad \ \leftarrow \text{Remainder}

Solution B: synthetic division

As you can see, all the math involved relates only to the coefficients of the variables and the constants. We could just as easily complete the division by omitting the variables, as long as we write the coefficients in the proper places.

Doing that, the problem looks like this: & \qquad \quad \ \ x^2 + 3x - 4 \qquad  \quad \leftarrow\text{Quotient}\\& 1-2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad 1-2\\& \qquad \ \  \searrow\\& \qquad \qquad +3 -10\\& \qquad \qquad \quad \ 3 -6\\& \qquad \qquad \quad \ \searrow\\& \qquad \qquad \qquad -4+13\\& \qquad \qquad \qquad \ -4+8\\& \qquad \qquad \qquad \quad \searrow\\& \qquad \qquad \qquad \qquad \ +5 \qquad \leftarrow \text{Remainder}

Since the underlined numerals are repetitions of those immediately above them, we can shorten the process by simply deleting them. Further, since these underlined numbers are products of the numbers in the quotient by the 1 in the divisor, we eliminate this 1.

Thus we get the following & \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad  -2\\& \qquad \ \searrow\\& \qquad \quad \ +3 -10\\& \qquad \qquad \quad \ -6\\& \qquad \qquad \ \searrow\\& \qquad \qquad \quad \ -4+13\\& \qquad \qquad \qquad \quad +8\\& \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}

It is also unnecessary to bring down the -10 and 13:

& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad  -2\\& \qquad \ \searrow\\& \qquad \quad \ +3\\& \qquad \qquad \quad \ -6\\& \qquad \qquad \ \searrow\\& \qquad \qquad \quad \ -4\\& \qquad \qquad \qquad \quad +8\\& \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}

Moving the numerals upward, we get

& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad \ \underline{ -2\ -6 \ +8\;\;}\\& \qquad \quad \ +3-4 \ +5 \qquad \ \leftarrow \text{Remainder}

When the numeral 1 (the first number in the quotient) is brought down to the last line, it will contain the remainder and the quotient,

& \qquad \quad -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \qquad \qquad\ \underline{ -2\ -6 \ +8\;\;}\\& \text{Quotient} \rightarrow 1 + 3 \ -4 \ +5 \qquad \ \leftarrow \text{Remainder}=+5

Therefore, the quotient is Q(x)=x^{2}+3x-4 and the remainder is R(x)=5 . This is synthetic division'.

Example B

Use synthetic division to find the quotient and the remainder of

\frac{3x^{3}-8x+1}{x+2}

Solution

First we write the divisor x+2 in the form x-c , as x-(-2) . Then use -2 as a “divisor” in the synthetic division as follows:

& \-2 \ \big ) \overline{3 \ \ \ 0 \ -8 \ \ \ \ 1}\\& \quad \ \underline{\downarrow -6 \ \ \ 12 -8}\\& \quad \  3 -6 \ \ \ 4 \ -7

Notice that 0 is used as the coefficient of the “missing” x^{2} term. Also, we wrote the coefficients of the dividend in descending order. The process of synthetic division is as follows: Bring down the first coefficient 3 and multiply by -2 (the divisor) to get -6, and then add 0 to -6 to get -6. Next, multiply -6 by -2 (the divisor) to get 12, and then add -8 to 12 to get 4. Finally, multiply 4 by -2 to get -8, and then add 1 to -8 to get the remainder, -7. As a result of this process, the quotient is

Q(x)=3x^{2}-6x+4

and the remainder is

R(x)=-7

In other words, since

f(x) & = D(x)Q(x)+R(x)\\& = (x+2)(3x^2-6x+4)+(-7)

Remember, this method will only work when the divisor is in the form x-c , that is, when the coefficient of x in the divisor is 1.

Example C

If h(x)=x^{3}-2x^{2}+5x-3 , evaluate

  • h(1)
  • h(-2)
  • h\left ( \frac{1}{2} \right )

Solution

We can simply use the synthetic division to evaluate h(x) at the given values. By the remainder theorem, the remainder is equal to h(c) .

  • Using synthetic division,

& \ 1 \ \big ) \overline{1  -2 \ \ \ 5  \ -3}\\& \quad \ \ \underline{\downarrow \ \ 1 -1 \ \ \ \ 4}\\& \quad \ \ 1 -1 \ \ \ 4 \ \ \ \ 1

Hence, h(1)=1 .

  • By synthetic division,

& \ -2 \ \big ) \overline{1  -2 \ \ \ 5  \ -3\;}\\& \qquad \quad \underline{\downarrow \ \ 2 \ \ \ 8 \ \ \ \ 26\;}\\& \qquad \quad 1 -4 \ 13 -29

Hence, h(-2)=-29 .

  • By synthetic division,

& \ \frac{1}{2} \ \big ) \overline{1  -2 \ \ \ 5  \ -3\;}\\& \qquad \underline{\downarrow \ \ \frac{1}{2} \  \frac{-3}{4} \ \ \frac{17}{8}\;}\\& \qquad 1 \ \ \frac{-3}{2} \ \frac{17}{4} \ \frac{-7}{8}

Hence, h\left ( \frac{1}{2} \right )=\frac{-7}{8} .

Vocabulary

Synthetic division is a concise 'shortcut' method of dividing polynomials.

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

A Dividend is the number (or polynomial) being divided, a divisor is the number (or polynomial) being divided 'into' the dividend, and a quotient is the result, occasionally with a remainder (the amount left over after all the even division has been completed).

Guided Practice

Problems

Divide using Long Division:

1) \frac{8x^3 - 7x^2 +10x - 5}{2x + 1}

2) \frac{x^3 +5x -4}{x^2 - x + 1}

Divide using Synthetic Division:

3) \frac{2x^4 + 5x^3 -2x -8}{x + 3}

Solutions

1) Solved in the video below:

- KhanAcademy: Algebraic Long Division

2) Solved in the video below:

- KhanAcademy: Dividing Polynomials with Remainders

3) Solved in the video below:

- clindelof: Synthetic Division One

Practice

Divide by using long division:

  1. (20x^2 - 13x + 2) divided by (4x - 1)
  2. (x^2 - 2x + 3) divided by (x + 5)
  3. \frac{y^4 - y^2 - 6y}{y^2 - 2}
  4. (x^3 + 2x^2 - x - 2) divided by (x + 2)
  5. \frac{x^4 - 1}{x^2 +1}

Divide using synthetic division:

  1. (7x^2 - 23x + 6) divided by (x - 3)
  2. (x^4 - 5x + 10) divided by (x + 3)
  3. (2x^2 + 13x - 8) divided by (x - \frac{1}{2})
  4. (x^4 + 6x^3 + 6x^2) divided by (x + 5)
  5. \frac{x^3 - 7x - 6}{x + 2}
  6. (8y^3 + y^4 + 16 + 32y + 24y^2) divided by (y + 2)

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

  1. P(x) = 2x^2 - 5x - 3 for x = 4
  2. P(x) = 4x^3 - 5x^2 + 3 for x = -1
  3. p(x) = 3x^3 - 5x^2 - x =2 for x = -\frac{1}{3}
  4. The area of a rectangle is  3x^3 - 11x^2 - 56x - 48 and the length is 3x + 4 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 V(h) = \frac {1}{4} \pi h^3 . The mass (in grams) of each sample in terms of height an be modeled by M(h) = \frac{1}{4}h^3 - h^2 + 5h . Write an expression that represents the density of the samples. (Hint: D = \frac{M}{V} )

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At Grade

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Date Created:

Nov 01, 2012

Last Modified:

Oct 28, 2014
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