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6.9: Long Division of Polynomials

Difficulty Level: At Grade Created by: CK-12
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The area of a rectangle is \begin{align*}6x^3 - 12x^2 + 4x - 8\end{align*}. The width of the rectangle is \begin{align*}2x - 4\end{align*}. What is the length?

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Khan Academy: Polynomial Division


Even though it does not seem like it, factoring is a form of division. Each factor goes into the larger polynomial evenly, without a remainder. For example, take the polynomial \begin{align*}2x^3-3x^2-8x+12\end{align*}. If we use factoring by grouping, we find that the factors are \begin{align*}(2x - 3)(x - 2)(x + 2)\end{align*}. If we multiply these three factors together, we will get the original polynomial. So, if we divide by \begin{align*}2x - 3\end{align*}, we should get \begin{align*}x^2 - 4\end{align*}.

\begin{align*}& \overset{\quad \ \ \ \quad {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{2x-3 \Big ) 2x^3-3x^2-8x+12}\end{align*}

How many times does \begin{align*}2x\end{align*} go into \begin{align*}2x^3\end{align*}? \begin{align*}x^2\end{align*} times.

\begin{align*}& \overset{\qquad \ \ \ \overset{{\color{blue}x^2}}{\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}} {{\color{red}2x}-3 \big ) {\color{red}2x^3}-3x^2-8x+12}\\ & \quad \quad \quad \underline{2x^3-3x^2 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;}\\ & \qquad \qquad \ \ 0\end{align*}

Place \begin{align*}x^2\end{align*} above the \begin{align*}x^2\end{align*} term in the polynomial.

Multiply \begin{align*}x^2\end{align*} by both terms in the divisor (\begin{align*}2x\end{align*} and -3) and place them until their like terms. Subtract from the dividend \begin{align*}(2x^3-3x^2-8x+12)\end{align*}. Pull down the next two terms and repeat.

\begin{align*}& \overset{\quad \quad \ \ \overset{\qquad \quad \ x^2 \qquad \ \ -4} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}2x}-3 \Big ) 2x^3-3x^2-8x+12}\\ & \quad \quad \quad \ \underline{2x^3-3x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad \ \quad \ \ {\color{red}-8x}+12\\ & \quad \quad \quad \quad \quad \quad \quad \quad \underline{-8x+12 \;\;}\end{align*}

\begin{align*}2x\end{align*} goes into \begin{align*}-8x\end{align*} 4 times.

After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtracting we notice that everything cancels out. Therefore, just like we thought, \begin{align*}x^2 - 4\end{align*} is a factor.

When dividing polynomials, not every divisor will go in evenly to the dividend. If there is a remainder, write it as a fraction over the divisor.

Example A

\begin{align*}(2x^3-6x^2+5x-20) \div (x^2-5)\end{align*}

Solution: Set up the problem using a long division bar.

\begin{align*}& \overset{\quad \ \ \ \quad {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2-5 \Big ) 2x^3-6x^2+5x-20}\end{align*}

How many times does \begin{align*}x^2\end{align*} go into \begin{align*}2x^3\end{align*}? \begin{align*}2x\end{align*} times.

\begin{align*}& \overset{\quad \ \ \ \quad \overset{\qquad \quad \ \ {\color{red}2x}} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}x^2}-5 \Big ) {\color{red}2x^3}-6x^2+5x-20}\\ & \quad \quad \quad \ \underline{2x^3-10x^2 \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad \quad 4x^2+5x-20\end{align*}

Multiply \begin{align*}2x\end{align*} by the divisor. Subtract that from the dividend.

Repeat the previous steps. Now, how many times does \begin{align*}x^2\end{align*} go into \begin{align*}4x^2\end{align*}? 4 times.

\begin{align*}& \overset{\quad \ \ \ \quad \overset{\qquad \qquad \ 2x+{\color{red}4}} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{{\color{red}x^2}-5 \Big ) 2x^3-6x^2+5x-20}\\ & \quad \quad \quad \ \underline{2x^3-10x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad \quad {\color{red}4x^2}+5x-20\\ & \quad \quad \quad \quad \quad \quad \quad \underline{4x^2 \qquad \ -20 \;}\\ &\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5x\end{align*}

At this point, we are done. \begin{align*}x^2\end{align*} cannot go into \begin{align*}5x\end{align*} because it has a higher degree. Therefore, \begin{align*}5x\end{align*} is a remainder. The complete answer would be \begin{align*}2x+4+\frac{5x}{x^2-5}\end{align*}.

Example B

\begin{align*}(3x^4+x^3-17x^2+19x-6) \div (x^2-2x+1)\end{align*}. Determine if \begin{align*}x^2-2x+1\end{align*} goes evenly into \begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}. If so, try to factor the divisor and quotient further.

Solution: First, do the long division. If \begin{align*}x^2-2x+1\end{align*} goes in evenly, then the remainder will be zero.

\begin{align*}& \overset{\qquad \qquad \quad \overset{\qquad \qquad 3x^2+7x-6} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2-2x+1 \Big ) 3x^4+x^3-17x^2+19x-6}\\ & \qquad \qquad \quad \ \ \underline{3x^4-6x^3+3x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \qquad \qquad \quad \ 7x^3-20x^2+19x\\ & \quad \quad \quad \qquad \qquad \quad \ \underline{7x^3-14x^2+7x\;\;\;\;\;\;\;\;}\\ &\quad \quad \quad \qquad \qquad \qquad \quad -6x^2+12x-6\\ &\quad \quad \quad \qquad \qquad \qquad \quad \ \underline{-6x^2+12x-6 \;\;\;}\\ &\quad \quad \quad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0\end{align*}

This means that \begin{align*}x^2-2x+1\end{align*} and \begin{align*}3x^2+7x-6\end{align*} both go evenly into \begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}. Let’s see if we can factor either \begin{align*}x^2-2x+1\end{align*} or \begin{align*}3x^2+7x-6\end{align*} further.

\begin{align*}x^2-2x+1=(x-1)(x-1)\end{align*} and \begin{align*}3x^2+7x-6=(3x-2)(x+3)\end{align*}.

Therefore, \begin{align*}3x^4+x^3-17x^2+19x-6=(x-1)(x-1)(x+3)(3x-2)\end{align*}. You can multiply these to check the work. A binomial with a degree of one is a factor of a larger polynomial, \begin{align*}f(x)\end{align*}, if it goes evenly into it. In this example, \begin{align*}(x-1)(x-1)(x+3)\end{align*} and \begin{align*}(3x-2)\end{align*} are all factors of \begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}. We can also say that 1, 1, -3, and \begin{align*}\frac{2}{3}\end{align*} are all solutions of \begin{align*}3x^4+x^3-17x^2+19x-6\end{align*}.

Factor Theorem: A polynomial, \begin{align*}f(x)\end{align*}, has a factor, \begin{align*}(x - k)\end{align*}, if and only if \begin{align*}f(k) = 0\end{align*}.

In other words, if \begin{align*}k\end{align*} is a solution or a zero, then the factor, \begin{align*}(x - k)\end{align*} divides evenly into \begin{align*}f(x)\end{align*}.

Example C

Determine if 5 is a solution of \begin{align*}x^3+6x^2-8x+15\end{align*}.

Solution: To see if 5 is a solution, we need to divide the factor into \begin{align*}x^3+6x^2-8x+15\end{align*}. The factor that corresponds with 5 is \begin{align*}(x - 5)\end{align*}.

\begin{align*}& \overset{\qquad \ \overset{\qquad \quad x^2+11x+5} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x-5 \Big ) x^3+6x^2-50x+15}\\ & \quad \quad \quad \ \underline{x^3-5x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad 11x^2-50x\\ & \quad \quad \quad \quad \quad \quad \underline{11x^2-55x \;\;\;\;\;\;}\\ &\quad \quad \quad \quad \quad \quad \quad \quad \quad 5x+15\\ &\quad \quad \quad \quad \quad \quad \quad \quad \quad \underline{5x-25 \;}\\ &\quad \quad \quad \quad \quad \quad \quad \quad \qquad \quad \ 40\end{align*}

Since there is a remainder, 5 is not a solution.

Intro Problem Revisit

First, do the long division.

\begin{align*}& \overset{\qquad \qquad \quad \overset{\qquad \qquad 3x^2 + 2} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{2x -4 \Big ) 6x^3-12x^2+4x-8}\\ & \qquad \qquad \quad \ \ \underline{6x^3-12x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \qquad \qquad \quad \ 4x - 8\\ & \quad \quad \quad \qquad \qquad \quad \ \underline{4x - 8\;\;\;\;\;\;\;\;}\\ &\quad \quad \quad \qquad \qquad \quad \ \ 0\end{align*}

This means that \begin{align*}2x - 4\end{align*} and \begin{align*}3x^2+2\end{align*} both go evenly into \begin{align*}6x^3-12x^2+4x-8\end{align*}.

\begin{align*}3x^2+2\end{align*} can't be factored further, so it is the rectangle's length.

Guided Practice

1. \begin{align*}(5x^4+6x^3-12x^2-3) \div (x^2+3)\end{align*}

2. Is \begin{align*}(x+4)\end{align*} a factor of \begin{align*}x^3-2x^2-51x-108\end{align*}? If so, find any other factors.

3. What are the real-number solutions to #2?

4. Determine if 6 is a solution to \begin{align*}2x^3-9x^2-12x-24\end{align*}.


1. Make sure to put a placeholder in for the \begin{align*}x-\end{align*}term.

\begin{align*}& \overset{\qquad \ \ \ \ \overset{\qquad \qquad 5x^2+6x-27} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x^2+3 \Big ) 5x^4+6x^3-12x^2+{\color{red}0x}-3}\\ & \quad \quad \quad \ \underline{5x^4 \qquad \ +15x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad 6x^3-27x^2+{\color{red}0x}-3\\ & \quad \quad \quad \quad \quad \quad \underline{6x^3 \quad \quad \quad \ +18x \;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \qquad \ \ -27x^2-18x-3\\ & \quad \quad \quad \quad \quad \qquad \quad \underline{-27x^2 \quad \quad \quad -81 \;\;}\\ & \quad \quad \quad \quad \quad \quad \qquad \qquad \quad -18x+78\end{align*}

The final answer is \begin{align*}5x^2+6x-27-\frac{18x-78}{x^2+3}\end{align*}.

2. Divide \begin{align*}(x + 4)\end{align*} into \begin{align*}x^3-2x^2-51x-108\end{align*} and if the remainder is zero, it is a factor.

\begin{align*}& \overset{\qquad \ \ \ \overset{\qquad \quad x^2-6x-27} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x+4 \Big ) x^3-2x^2-51x-108}\\ & \quad \quad \ \ \underline{x^3+4x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \ -6x^2-51x\\ & \quad \quad \quad \quad \ \ \underline{-6x^2-24x \;\;\;\;\;\;\;\;\;\;}\\ &\quad \quad \quad \quad \quad \quad \quad -27x-108\\ &\quad \quad \quad \quad \quad \quad \quad \ \underline{-27x-108 \;}\\ &\quad \quad \quad \quad \quad \quad \qquad \qquad \qquad 0\end{align*}

\begin{align*}x + 4\end{align*} is a factor. Let’s see if \begin{align*}x^2 - 6x -27\end{align*} factors further. Yes, the factors of -27 that add up to -6 are -9 and 3. Therefore, the factors of \begin{align*}x^3-2x^2-51x-108\end{align*} are \begin{align*}(x + 4), (x - 9)\end{align*}, and \begin{align*}(x + 3)\end{align*}.

3. The solutions would be -4, 9, and 3; the opposite sign of each factor.

4. To see if 6 is a solution, we need to divide \begin{align*}(x - 6)\end{align*} into \begin{align*}2x^3-9x^2-12x-24\end{align*}.

\begin{align*}& \overset{\qquad \ \ \overset{\qquad \quad \ 2x^2+3x+6} {\underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}}{x-6 \Big ) 2x^3-9x^2-12x-24}\\ & \quad \quad \ \ \underline{2x^3-12x^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & \quad \quad \quad \quad \quad \quad 3x^2-12x\\ & \quad \quad \quad \quad \quad \quad \underline{3x^2-18x \;\;\;\;\;\;\;\;\;\;}\\ &\quad \quad \quad \quad \quad \quad \quad \quad 6x-24\\ &\quad \quad \quad \quad \quad \quad \quad \quad \underline{6x-36 \;\;\;\;\;\;}\\ &\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ 12\end{align*}

Because the remainder is not zero, 6 is not a solution.


Long division (of polynomial)
The process of dividing polynomials where the divisor has two or more terms.
The polynomial that divides into another polynomial.
The polynomial that the divisor goes into. The polynomial under the division bar.
The answer to a division problem.


Divide the following polynomials using long division.

  1. \begin{align*}(2x^3+5x^2-7x-6) \div (x+1)\end{align*}
  2. \begin{align*}(x^4-10x^3+15x-30) \div (x-5)\end{align*}
  3. \begin{align*}(2x^4-8x^3+4x^2-11x-1) \div (x^2-1)\end{align*}
  4. \begin{align*}(3x^3-4x^2+5x-2) \div (3x+2)\end{align*}
  5. \begin{align*}(3x^4-5x^3-21x^2-30x+8) \div (x-4)\end{align*}
  6. \begin{align*}(2x^5-5x^3+6x^2-15x+20) \div (2x^2+3)\end{align*}

Determine all the real-number solutions to the following polynomials, given one factor.

  1. \begin{align*}x^3-9x^2+27x-15; (x+5)\end{align*}
  2. \begin{align*}x^3+4x^2-9x-36; (x+4)\end{align*}
  3. \begin{align*}2x^3+7x^2-7x-30; (x-2)\end{align*}
  4. Determine all the real number solutions to the following polynomials, given one zero.
  5. \begin{align*}6x^3-37x^2+5x+6; 6\end{align*}
  6. \begin{align*}6x^3-41x^2+58x+15; 5\end{align*}
  7. \begin{align*}x^3+x^2-16x-16; 4 \end{align*}
  8. Find the equation of a polynomial with the given zeros.
  9. 4, -2, and \begin{align*}\frac{3}{2}\end{align*}
  10. 1, 0, and 3
  11. -5, -1, and \begin{align*} \frac{3}{4}\end{align*}
  12. Challenge Find two polynomials with the zeros 8, 5, 1, and -1.

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Long division (of polynomial)

Long division of polynomials is the process of dividing polynomials when the divisor has two or more terms.


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

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