College Algebra : Dividing Polynomials

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #1 : Simplifying Polynomials

Divide the trinomial below by \(\displaystyle 9x\).

\(\displaystyle 18x^{2}+ 9x -3\)

Possible Answers:

\(\displaystyle 2x+1-\frac{1}{3x}\)

\(\displaystyle 2x^{2}+ x -\frac{1}{3}\)

\(\displaystyle -3x+1+\frac{1}{2x}\)

\(\displaystyle 2x+1-\frac{3}{x}\)

\(\displaystyle -x+1\)

Correct answer:

\(\displaystyle 2x+1-\frac{1}{3x}\)

Explanation:

\(\displaystyle 18x^{2}+ 9x -3\)

We can accomplish this division by re-writing the problem as a fraction.

\(\displaystyle \frac{18x^2+9x-3}{9x}\)

The denominator will distribute, allowing us to address each element separately.

\(\displaystyle \frac{18x^{2}+ 9x -3}{9x}=\frac{18x^{2}}{9x}+\frac{ 9x}{9x}-\frac{3}{9x}\)

Now we can cancel common factors to find our answer.

\(\displaystyle (\frac{18}{9}* \frac{x^{2}}{x})+1-(\frac{3}{9}* \frac{1}{x})\)

\(\displaystyle 2x+1-\frac{1}{3x}\)

Example Question #1 : Dividing Polynomials

Simplify:

\(\displaystyle y=\frac{(x^2-16)(x+7)}{x+4}+(x-1)^2+13\)

Possible Answers:

\(\displaystyle y=2x^2+x-14\)

\(\displaystyle y=\frac{(x^2-16)(x+7)}{x+4}+x^2-2x+14\)

\(\displaystyle y=2(x^2+x-7)\)

\(\displaystyle y=x^2-x-14\)

\(\displaystyle y=\frac{x^3+7x^2-16x-112}{x+4}+(x-1)^2+13\)

Correct answer:

\(\displaystyle y=2x^2+x-14\)

Explanation:

\(\displaystyle y=\frac{(x^2-16)(x+7)}{x+4}+(x-1)^2+13\)

First, factor the numerator of the quotient term by recognizing the difference of squares:

\(\displaystyle y=\frac{(x-4)(x+4)(x+7)}{x+4}+(x-1)^2+13\)

Cancel out the common term from the numerator and denominator:

\(\displaystyle y=(x-4)(x+7)+(x-1)^2+13\)

FOIL (First Outer Inner Last) the first two terms of the equation:

\(\displaystyle y=x^2+7x-4x-28+x^2-2x+1+13\)

Combine like terms:

\(\displaystyle y=2x^2+x-14\)

Example Question #2 : Dividing Polynomials

Divide:

\(\displaystyle (x^{3} -23x+10) \div (x-5)\)

Possible Answers:

\(\displaystyle x^{2} + 5x + 2\)

\(\displaystyle x^{2} - 5x + 16 + \frac{87}{x-5}\)

\(\displaystyle x^{2} + 5x + 2 + \frac{20}{x-5}\)

\(\displaystyle x^{2}+1+\frac{12}{x-5}\)

\(\displaystyle x^{2} - 5x + 16 - \frac{73}{x-5}\)

Correct answer:

\(\displaystyle x^{2} + 5x + 2 + \frac{20}{x-5}\)

Explanation:

First, rewrite this problem so that the missing \(\displaystyle x^{2}\) term is replaced by \(\displaystyle 0x^{2}\)

\(\displaystyle (x^{3} + 0x^{2} -23x+10) \div (x-5)\)

Divide the leading coefficients:

\(\displaystyle x^{3} \div x = x^{2}\), the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

\(\displaystyle x^{2} (x-5) = x^{3} -5x^{2}\)

\(\displaystyle (x^{3} + 0x^{2} -23x+10) - (x^{3} -5x^{2}) = 5x^{2}-23x+10\)

Repeat this process with each difference:

\(\displaystyle 5x^{2} \div x = 5x\), the second term of the quotient

\(\displaystyle 5x (x-5) = 5 x^{2} -25\)

\(\displaystyle 5x^{2}-23x+10 - ( 5x^{2} -25x) = 2x+10\)

One more time:

\(\displaystyle 2x \div x = 2\), the third term of the quotient

\(\displaystyle 2 (x-5) = 2x-10\)

\(\displaystyle 2x+10 - (2x-10) = 20\), the remainder

The quotient is \(\displaystyle x^{2} + 5x + 2\) and the remainder is \(\displaystyle 20\); this can be rewritten as a quotient of 

\(\displaystyle x^{2} + 5x + 2 + \frac{20}{x-5}\)

Example Question #3 : Dividing Polynomials

Divide:

\(\displaystyle \left (x^{2} + 6x -7 \right )\div (x+3)\)

 

Possible Answers:

\(\displaystyle x+ 3 + \frac{2}{x+3}\)

\(\displaystyle x+ 3 - \frac{16}{x+3}\)

\(\displaystyle x- 3 - \frac{16}{x+3}\)

\(\displaystyle x- 3 + \frac{2}{x+3}\)

\(\displaystyle x+ 3 + \frac{16}{x+3}\)

Correct answer:

\(\displaystyle x+ 3 - \frac{16}{x+3}\)

Explanation:

Divide the leading coefficients to get the first term of the quotient:

\(\displaystyle x^{2} \div x = x\), the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

\(\displaystyle x (x+3) = x^{2} + 3x\)

\(\displaystyle \left (x^{2} + 6x -7 \right ) - (x^{2}+3x) = 3x -7\)

Repeat these steps with the differences until the difference is an integer. As it turns out, we need to repeat only once:

\(\displaystyle 3x\div x = 3\), the second term of the quotient

\(\displaystyle 3 (x+3) = 3x + 9\)

\(\displaystyle (3x -7) - (3x + 9) = -16\), the remainder

 

Putting it all together, the quotient can be written as \(\displaystyle x+ 3 - \frac{16}{x+3}\).

Example Question #1 : Polynomial Functions

Simplify the following expression:

\(\displaystyle \frac{(x-25)}{(x^2-625)}\)

Possible Answers:

\(\displaystyle \frac{1}{x+25}\)

\(\displaystyle \frac{1}{x^2+25}\)

\(\displaystyle {x+25}\)

\(\displaystyle \frac{1}{x-25}\)

Correct answer:

\(\displaystyle \frac{1}{x+25}\)

Explanation:

Simplify the following expression:

\(\displaystyle \frac{(x-25)}{(x^2-625)}\)

To begin, we need to recognize the bottom as a difference of squares. Rewrite it as follows.

\(\displaystyle \frac{(x-25)}{(x+25)(x-25)}=\frac{1}{x+25}\)

So our answer is:

\(\displaystyle \frac{1}{x+25}\)

Example Question #3 : Dividing Polynomials

Simplify the following expression:

\(\displaystyle \frac{(3x)(3x^2+5x)}{x}\)

Possible Answers:

\(\displaystyle 9x^2+15x\)

\(\displaystyle 3x^2+5x\)

\(\displaystyle 9x^3+15x^2\)

\(\displaystyle 9x+15\)

Correct answer:

\(\displaystyle 9x^2+15x\)

Explanation:

Simplify the following expression:

\(\displaystyle \frac{(3x)(3x^2+5x)}{x}\)

First, let's multiply the 3x through:

\(\displaystyle \frac{(3x)(3x^2+5x)}{x}=\frac{9x^3+15x^2}{x}\)

Next, divide out the x from the bottom:

\(\displaystyle \frac{9x^3+15x^2}{x}=9x^2+15x\)

So our answer is:

\(\displaystyle 9x^2+15x\)

Example Question #4 : Polynomial Functions

\(\displaystyle \begin{align*}&\text{Simplify the expresssion:}\\&\frac{x^{2}+30x+225}{x+15}\end{align*}\)

Possible Answers:

\(\displaystyle x+15\)

\(\displaystyle x+5\)

\(\displaystyle x+23\)

\(\displaystyle x+3\)

Correct answer:

\(\displaystyle x+15\)

Explanation:

\(\displaystyle \begin{align*}&\textbf{To perform a polynomial long division, it is often}\\&\textbf{helpful to space terms into columns ordered by power.}\\&\textbf{This includes zero terms. For example: }x^2+0x+5\\&\textbf{When performing each division step, choose a coefficient}\\&\textbf{Which eliminates the highest ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+15}\\x&&+15|x^{2}&&+30x&&+225\\&&-(x^{2}&&+15x)\\&&&&15x&&+225\\&&&&-(15x&&+225)\\\end{align*}\)

Example Question #1 : Polynomial Functions

\(\displaystyle \begin{align*}&\text{Simplify the expresssion:}\\&\frac{x^{2}+x-30}{x-5}\end{align*}\)

Possible Answers:

\(\displaystyle x-26\)

\(\displaystyle x-6\)

\(\displaystyle x+1\)

\(\displaystyle x+6\)

Correct answer:

\(\displaystyle x+6\)

Explanation:

\(\displaystyle \begin{align*}&\textbf{To perform a polynomial long division, it is often}\\&\textbf{helpful to space terms into columns ordered by power.}\\&\textbf{This includes zero terms. For example: }x^2+0x+5\\&\textbf{When performing each division step, choose a coefficient}\\&\textbf{Which eliminates the highest ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+6}\\x&&-5|x^{2}&&+x&&-30\\&&-(x^{2}&&-5x)\\&&&&6x&&-30\\&&&&-(6x&&-30)\\\end{align*}\)

Example Question #5 : Dividing Polynomials

\(\displaystyle \begin{align*}&\text{Simplify the expresssion:}\\&\frac{x^{2}+24x+143}{x+13}\end{align*}\)

Possible Answers:

\(\displaystyle x-5\)

\(\displaystyle x+25\)

\(\displaystyle x-27\)

\(\displaystyle x+11\)

Correct answer:

\(\displaystyle x+11\)

Explanation:

\(\displaystyle \begin{align*}&\textbf{To perform a polynomial}\\&\textbf{long division, it is often}\\&\textbf{helpful to space terms}\\&\textbf{into columns ordered by power.}\\&\textbf{This includes zero terms.}\\&\textbf{For example: }x^2+0x+5\\&\textbf{When performing each}\\&\textbf{division step, choose a coefficient}\\&\textbf{Which eliminates the highest}\\&\textbf{ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+11}\\x&&+13|x^{2}&&+24x&&+143\\&&-(x^{2}&&+13x)\\&&&&11x&&+143\\&&&&-(11x&&+143)\\\end{align*}\)

Example Question #4 : Dividing Polynomials

\(\displaystyle \begin{align*}&\text{Simplify the expresssion:}\\&\frac{x^{5}+x^{4}-353x^{3}-641x^{2}+24832x+98560}{x^{2}-9x-112}\end{align*}\)

Possible Answers:

\(\displaystyle x^{3}+28x^{2}+5x-27\)

\(\displaystyle x^{3}+10x^{2}-151x-880\)

\(\displaystyle x^{3}-7x^{2}-16x-6\)

\(\displaystyle x^{3}-25x^{2}-22x+27\)

Correct answer:

\(\displaystyle x^{3}+10x^{2}-151x-880\)

Explanation:

\(\displaystyle \begin{align*}&\textbf{To perform a polynomial}\\&\textbf{long division, it is often}\\&\textbf{helpful to space terms}\\&\textbf{into columns ordered by power.}\\&\textbf{This includes zero terms.}\\&\textbf{For example: }x^2+0x+5\\&\textbf{When performing each}\\&\textbf{division step, choose a coefficient}\\&\textbf{Which eliminates the highest}\\&\textbf{ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&&&&&\mathbf{x^{3}}&&\mathbf{+10x^{2}}&&\mathbf{-151x}&&\mathbf{-880}\\x^{2}&&-9x&&-112|x^{5}&&+x^{4}&&-353x^{3}&&-641x^{2}&&+24832x&&+98560\\&&&&-(x^{5}&&-9x^{4}&&-112x^{3})\\&&&&&&10x^{4}&&-241x^{3}&&-641x^{2}\\&&&&&&-(10x^{4}&&-90x^{3}&&-1120x^{2})\\&&&&&&&&-151x^{3}&&479x^{2}&&24832x\\&&&&&&&&-(-151x^{3}&&1359x^{2}&&16912x)\\&&&&&&&&&&-880x^{2}&&7920x&&98560\\&&&&&&&&&&-(-880x^{2}&&7920x&&98560)\\\end{align*}\)

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