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

### 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 $2x^3-9x^2-12x-24$ .

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.

### Vocabulary Language: English

Long division (of polynomial)

Long division (of polynomial)

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

Quotient

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