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Long Division of Polynomials

Identify factors of polynomials using long division

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Long Division of Polynomials

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?

Long Division of Polynomials

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*}\require{enclose} 2x-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12} \end{align*}

How many times does \begin{align*}2x\end{align*} go into \begin{align*}2x^3?\end{align*} Since \begin{align*}2x(x^2)=2x^3,\end{align*} it goes in \begin{align*}x^2\end{align*} times.

\begin{align*}\require{enclose} {x}^{2}\phantom{00000000000000}\\ 2x-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12}\\ \underline{2{x}^{3}-3{x}^{2}}\phantom{000000000}\\ 0\phantom{0000000000} \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 under 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*}\require{enclose} {x}^{2}\phantom{00000}-4\phantom{00000}\\ {\color{red}{2x}}-3\enclose{longdiv}{2{x}^{3}-3{x}^{2}-8x+12}\\ \underline{2{x}^{3}-3{x}^{2}\quad\quad\quad\phantom{00}}\\ {\color{red}-8x}+12\\ \underline{-8x+12}\end{align*}

Since \begin{align*}2x(\text{-}4)=\text{-}8x,\end{align*} \begin{align*}2x\end{align*} goes into \begin{align*}-8x\end{align*} a total of -4 times.

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

Let's divide the following polynomials using long division.

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

Set up the problem using a long division bar.

\begin{align*}\require{enclose} {x}^{2}-5\enclose{longdiv}{2{x}^{3}-6{x}^{2}+5x-20}\end{align*}

How many times does \begin{align*}x^2\end{align*} go into \begin{align*}2x^3?\end{align*} Since \begin{align*}x^2(2x)=2x^3,\end{align*} it goes in \begin{align*}2x\end{align*} times.

\begin{align*}\require{enclose} \color{red}2x\phantom{00000000000000}\\ {\color{red}{x}^{2}}-5\enclose{longdiv}{{\color{red}2{x}^{3}}-6{x}^{2}+5x-20}\\ \underline{2{x}^{3}-10{x}^{2}\quad\quad\quad\phantom{0}}\\ 4{x}^{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*}? It goes in 4 times.

\begin{align*}\require{enclose} 2x\phantom{00}\color{red}+4\phantom{000000000}\\ {\color{red}{x}^{2}}-5\enclose{longdiv}{2{x}^{3}-6{x}^{2}+5x-20}\\ \underline{2{x}^{3}-10{x}^{2}\quad\quad\quad\phantom{0}}\\ {\color{red}4{x}^{2}}+5x-20\\ \underline{4{x}^{2}\phantom{0000}-20}\\ 5x\quad\phantom{00}\end{align*}

This is the limit of this process. \begin{align*}x^2\end{align*} cannot go evenly 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*}

  1. \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.

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*}\require{enclose} 3{x}^{2}+7x\phantom{000}-6\phantom{000000000}\\ {x}^{2}-2x+1\enclose{longdiv}{3{x}^{4}+{x}^{3}-17{x}^{2}+19x-6}\\ \underline{3{x}^{4}-6{x}^{3}\phantom{00}3{x}^{2}\phantom{000000000}}\\ 7{x}^{3}-20{x}^{2}+19x\phantom{000}\\ \underline{7{x}^{3}-14{x}^{2}+\phantom{0}7x\phantom{000}}\\ -6{x}^{2}+12x-6\\ \underline{-6{x}^{2}+12x-6}\\ 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 problem, \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*}. This indicates 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*}.

Now, let's determine if 5 is a solution of \begin{align*}x^3+6x^2-8x+15\end{align*}.

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*}\require{enclose} {x}^{2}+11x\phantom{0}+5\phantom{00000}\\ x-5\enclose{longdiv}{{x}^{3}+6{x}^{2}-50x+15}\\ \underline{{x}^{3}-\phantom{0}5{x}^{2}\phantom{000000000}}\\ 11{x}^{2}-\phantom{0}50x\phantom{0000}\\ \underline{11{x}^{2}-\phantom{0}55x\phantom{0000}}\\ 5x+15\\ \underline{5x-25}\\ 40\end{align*}

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


Example 1

Earlier, you were asked to find the length of the rectangle. 

First, do the long division.

\begin{align*}\require{enclose} 3{x}^{2}\phantom{0000000}+2\phantom{0000}\\ 2x-4\enclose{longdiv}{6{x}^{3}-12{x}^{2}+4x-8}\\ \underline{6{x}^{3}-12{x}^{2}\quad\quad\quad\phantom{0}}\\ 4x-8\\ \underline{4x-8}\\ 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.

Example 2

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

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

\begin{align*}\require{enclose} 5{x}^{2}+6x\phantom{00}-27\phantom{000000000}\\ {x}^{2}+3\enclose{longdiv}{5{x}^{4}+6{x}^{3}-12{x}^{2}+{\color{red}0x}-3}\\ \underline{5{x}^{4}\phantom{00000}+15{x}^{2}\phantom{00000000}}\\ 6{x}^{3}-27{x}^{2}+{\color{red}0x}\quad\phantom{0}\\ \underline{6{x}^{3}\quad\phantom{000}+18x\phantom{000}}\\ -27{x}^{2}-18x-3\\ \underline{-27{x}^{2}\quad\phantom{00}-81}\\ -18x+78\end{align*}

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

Example 3

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.

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*}\require{enclose} {x}^{2}-6x-27\phantom{0000000}\\ x+4\enclose{longdiv}{{x}^{3}-2{x}^{2}-51x-108}\\ \underline{{x}^{3}+4{x}^{2}\phantom{00000000000}}\\ -6{x}^{2}-51x\phantom{000000}\\ \underline{-6{x}^{2}-24x\phantom{000000}}\\ -27x-108\\ \underline{-27x-108}\\ 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*}.

Example 4

What are the real-number solutions to Example 3?

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

Example 5

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

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*}\require{enclose} 2{x}^{2}+3x\phantom{0}+6\phantom{0}\phantom{00000}\\ x-6\enclose{longdiv}{2{x}^{3}-9{x}^{2}-12x-24}\\ \underline{2{x}^{3}-12{x}^{2}\phantom{000000000}}\\ 3{x}^{2}-12x\phantom{0000}\\ \underline{3{x}^{2}-18x\phantom{0000}}\\ 6x-24\\ \underline{6x-36}\\ 12\end{align*}

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


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*}

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

  1. \begin{align*}6x^3-37x^2+5x+6; 6\end{align*}
  2. \begin{align*}6x^3-41x^2+58x-15; 5\end{align*}
  3. \begin{align*}x^3+x^2-16x-16; 4 \end{align*}

Find the equation of a polynomial with the given zeros.

  1. 4, -2, and \begin{align*}\frac{3}{2}\end{align*}
  2. 1, 0, and 3
  3. -5, -1, and \begin{align*} \frac{3}{4}\end{align*}
  4. Challenge Find two polynomials with the zeros 8, 5, 1, and -1.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 6.9. 

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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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