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Long Division of Polynomials

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Long Division of Polynomials

The area of a rectangle is 6x^3 - 12x^2 + 4x - 8 . The width of the rectangle is 2x - 4 . What is the length?

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Khan Academy: Polynomial Division

Guidance

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 2x^3-3x^2-8x+12 . If we use factoring by grouping, we find that the factors are (2x - 3)(x - 2)(x + 2) . If we multiply these three factors together, we will get the original polynomial. So, if we divide by 2x - 3 , we should get x^2 - 4 .

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}      &         &          &          &      &       \\ \cline{2-6} \rule{0pt}{2.3ex}2x-3~ & \Huge)~ &   2x^3   &\,-3 x^2  &\,-8x &\, +12 \\\end{array}

How many times does 2x go into 2x^3 ? x^2 times.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrr}                   &         &\, {\color{blue}x^2} &          &   \\ \cline{2-5} \rule{0pt}{2.3ex}{\color{red}2x}-3~ & \Huge)~ & {\color{red}2x^3}   &\,-3 x^2  &\,-8x + 12 \\                   &         &   2x^3 &\,-3 x^2    &   \\ \cline{3-5} \rule{0pt}{2.5ex}                   &         &        &\, 0~~      & \end{array}

Place x^2 above the x^2 term in the polynomial.

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

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}                   &         &\, x^2 &         &   -4  &       \\ \cline{2-6} \rule{0pt}{2.3ex}{\color{red}2x}-3~ & \Huge)~ &  2x^3 &\,-3 x^2 &\, -8x &\, +12 \\                   &         &  2x^3 &\,-3 x^2 &               \\ \cline{3-6} \rule{0pt}{2.5ex}                   &         &       &  &\,{\color{red}-8x} &\, +12 \\                    &         &       &         &\, -8x &\, +12 \\ \cline{5-6}\end{array}

2x goes into -8x 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, x^2 - 4 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

(2x^3-6x^2+5x-20) \div (x^2-5)

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

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}       &         &       &         &       &       \\ \cline{2-6} \rule{0pt}{2.3ex}x^2-5~ & \Huge)~ &  2x^3 &\, -6x^2 &\, +5x &\, -20 \\\end{array}

How many times does x^2 go into 2x^3 ? 2x times.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}                    &         &\, {\color{red}2x}  &         &       &       \\ \cline{2-6} \rule{0pt}{2.3ex}{\color{red}x^2}-5~ & \Huge)~ &  {\color{red}2x^3} &\, -6x^2 &\, +5x &\, -20 \\                    &         &             2x^3   &\,-10x^2 &       &       \\ \cline{3-6} \rule{0pt}{2.5ex}                    &         &                    &\,  4x^2 &\, +5x &\, -20\end{array}

Multiply 2x by the divisor. Subtract that from the dividend.

Repeat the previous steps. Now, how many times does x^2 go into 4x^2 ? 4 times.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}                    &         &   2x &\, {\color{red}+4} &   &   \\ \cline{2-6} \rule{0pt}{2.3ex}{\color{red}x^2}-5~ & \Huge)~ & 2x^3 &\, -6x^2       &\, +5x &\, -20 \\                    &         & 2x^3 &\,-10x^2       &       &       \\ \cline{3-6} \rule{0pt}{2.5ex}                    &         &   &\, {\color{red}4x^2} &\, +5x &\, -20 \\                    &         &      &\,  4x^2       &\,     &\, -20 \\ \cline{4-6} \rule{0pt}{2.5ex}                    &         &      &               &\,  5x &\,   \\\end{array}

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

Example B

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

Solution: First, do the long division. If x^2-2x+1 goes in evenly, then the remainder will be zero.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrrr}          &         & 3x^2 &\,  +7x  &\,  -6   &        &      \\ \cline{2-7} \rule{0pt}{2.3ex}x^2-2x+1~ & \Huge)~ & 3x^4 &\, +x^3  &\,-17x^2 &\, +19x &\, -6 \\          &         & 3x^4 &\, -6x^3 &\,  3x^2 &        &      \\ \cline{3-7} \rule{0pt}{2.5ex}          &         &      &\,  7x^3 &\,-20x^2 &\, +19x &\,    \\          &         &      &\,  7x^3 &\,-14x^2 &\,  +7x &\,    \\ \cline{4-7} \rule{0pt}{2.5ex}          &         &      &\,       &\, -6x^2 &\, +12x &\, -6 \\          &         &      &\,       &\, -6x^2 &\, +12x &\, -6 \\ \cline{5-7} \rule{0pt}{2.5ex}          &         &      &\,       &         &       &\,  0  \\\end{array}

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

x^2-2x+1=(x-1)(x-1) and 3x^2+7x-6=(3x-2)(x+3) .

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

Factor Theorem: A polynomial, f(x) , has a factor, (x - k) , if and only if f(k) = 0 .

In other words, if k is a solution or a zero , then the factor, (x - k) divides evenly into f(x) .

Example C

Determine if 5 is a solution of x^3+6x^2-8x+15 .

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

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}     &         &\, x^2 &\, +11x  &\,  +5  &       \\ \cline{2-6} \rule{0pt}{2.3ex}x-5~ & \Huge)~ &\, x^3 &\, +6x^2 &\, -50x &\, +15 \\     &         &\, x^3 &\, -5x^2 &        &       \\ \cline{3-6} \rule{0pt}{2.5ex}     &         &       &\, 11x^2 &\, -50x &\,     \\     &         &       &\, 11x^2 &\, -55x &\,     \\ \cline{4-6} \rule{0pt}{2.5ex}     &         &       &         &\,   5x &\, +15 \\     &         &       &         &\,   5x &\, -25 \\ \cline{5-6} \rule{0pt}{2.5ex}     &         &       &         &\,      &\,  40 \\\end{array}

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

Intro Problem Revisit

First, do the long division.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}      &         &\, 3x^2 &         &   +2  &       \\ \cline{2-6} \rule{0pt}{2.3ex}2x-4~ & \Huge)~ &\, 6x^3 &\,-12x^2 &\, +4x &\,  -8 \\      &         &\, 6x^3 &\,-12x^2 &       &       \\ \cline{3-6} \rule{0pt}{2.5ex}      &         &        &         &\,  4x &\,  -8 \\       &         &        &         &\,  4x &\,  -8 \\ \cline{5-6}      &         &        &         &\,     &\,   0 \\ \end{array}

This means that 2x - 4 and 3x^2+2 both go evenly into 6x^3-12x^2+4x-8 .

3x^2+2 can't be factored further, so it is the rectangle's length.

Guided Practice

1. (5x^4+6x^3-12x^2-3) \div (x^2+3)

2. Is (x+4) a factor of x^3-2x^2-51x-108 ? If so, find any other factors.

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

4. Determine if 6 is a solution to 2x^3-9x^2-12x-24 .

Answers

1. Make sure to put a placeholder in for the x- term.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrrr}       &         &\, 5x^2 &\,  +6x  &\,  -27   &                 &      \\ \cline{2-7} \rule{0pt}{2.3ex}x^2+3~ & \Huge)~ &\, 5x^4 &\, +6x^3 &\, -12x^2 &\,{\color{red}+0x} &\, -3 \\       &         &\, 5x^4 &\,       &\, +15x^2 &                 &      \\ \cline{3-7} \rule{0pt}{2.5ex}       &         &        &\,  6x^3 &\, -27x^2 &\,{\color{red}+0x} &      \\       &         &        &\,  6x^3 &\,        &\,   +18x        &      \\ \cline{4-7} \rule{0pt}{2.5ex}       &         &        &         &\, -27x^2 &\,   -18x        &\, -3   \\       &         &        &         &\, -27x^2 &\,               &\, -81  \\ \cline{5-7} \rule{0pt}{2.5ex}       &         &        &         &          &\,   -18x        &\, +78  \\\end{array}

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

2. Divide (x + 4) into x^3-2x^2-51x-108 and if the remainder is zero, it is a factor.

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}     &         &\, x^2 &\,  -6x  &\, -27  &        \\ \cline{2-6} \rule{0pt}{2.3ex}x+4~ & \Huge)~ &\, x^3 &\, -2x^2 &\, -51x &\, -108 \\     &         &\, x^3 &\, +4x^2 &        &        \\ \cline{3-6} \rule{0pt}{2.5ex}     &         &       &\, -6x^2 &\, -51x &        \\     &         &       &\, -6x^2 &\, -24x &        \\ \cline{4-6} \rule{0pt}{2.5ex}     &         &       &         &\, -27x &\, -108 \\     &         &       &         &\, -27x &\, -108 \\ \cline{5-6} \rule{0pt}{2.5ex}     &         &       &         &        &      0 \\\end{array}

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

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 (x - 6) into 2x^3-9x^2-12x-24 .

\renewcommand{\arraycolsep}{0pt}\begin{array}{clrrrr}     &         &\, 2x^2 &\,   +3x  &\,  +6  &       \\ \cline{2-6} \rule{0pt}{2.3ex}x-6~ & \Huge)~ &\, 2x^3 &\,  -9x^2 &\, -12x &\, -24 \\     &         &\, 2x^3 &\, -12x^2 &        &       \\ \cline{3-6} \rule{0pt}{2.5ex}     &         &        &\,   3x^2 &\, -12x &       \\     &         &        &\,   3x^2 &\, -18x &       \\ \cline{4-6} \rule{0pt}{2.5ex}     &         &        &          &\,   6x &\, -24 \\     &         &        &          &\,   6x &\, -36 \\ \cline{5-6} \rule{0pt}{2.5ex}     &         &        &          &        &    12 \\\end{array}

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

Vocabulary

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

Practice

Divide the following polynomials using long division.

  1. (2x^3+5x^2-7x-6) \div (x+1)
  2. (x^4-10x^3+15x-30) \div (x-5)
  3. (2x^4-8x^3+4x^2-11x-1) \div (x^2-1)
  4. (3x^3-4x^2+5x-2) \div (3x+2)
  5. (3x^4-5x^3-21x^2-30x+8) \div (x-4)
  6. (2x^5-5x^3+6x^2-15x+20) \div (2x^2+3)

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

  1. x^3-9x^2+27x-15; (x+5)
  2. x^3+4x^2-9x-36; (x+4)
  3. 2x^3+7x^2-7x-30; (x-2)

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

  1. 6x^3-37x^2+5x+6; 6
  2. 6x^3-41x^2+58x-15; 5
  3. x^3+x^2-16x-16; 4

Find the equation of a polynomial with the given zeros.

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

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