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Common Core: High School - Algebra : Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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Example Questions

Example Question #1 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ x^{3} - 8 x^{2} + 6 x + 6 }{ x }$

Possible Answers:

$x^{2} - 8 x + 6$

$x^{2} - 8 x + 6 + \frac{6}{x}$

$\frac{6}{x}$

Correct answer:

$x^{2} - 8 x + 6 + \frac{6}{x}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & ~ & ~\\ \hline ~ & 1 & -8 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & 0 & ~\\ \hline ~ & 1 & -8 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & 0 & ~\\ \hline ~ & 1 & -8 & 6 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & 0 & 0 \\ \hline ~ & 1 & -8 & 6 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 1 & -8 & 6 & 6 \\ ~ & ~ & 0 & 0 & 0 \\ \hline ~ & 1 & -8 & 6 & 6 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = x^{2} - 8 x + 6$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = 6$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = x^{2} - 8 x + 6 + \frac{6}{x}$

Report an Error

Example Question #1 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ 7 x^{3} - 4 x^{2} + 2 x + 4 }{ x - 4 }$

Possible Answers:

$7 x^{2} + 24 x + 98 + \frac{396}{x - 4}$

525

396

$7 x^{2} + 24 x + 98$

$\frac{396}{x - 4}$

Correct answer:

$7 x^{2} + 24 x + 98 + \frac{396}{x - 4}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & ~ & ~\\ \hline ~ & 7 & 24 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & 96 & ~\\ \hline ~ & 7 & 24 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & 96 & ~\\ \hline ~ & 7 & 24 & 98 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & 96 & 392 \\ \hline ~ & 7 & 24 & 98 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 4 & 7 & -4 & 2 & 4 \\ ~ & ~ & 28 & 96 & 392 \\ \hline ~ & 7 & 24 & 98 & 396 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = 7 x^{2} + 24 x + 98$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = 396$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x - 4$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = 7 x^{2} + 24 x + 98 + \frac{396}{x - 4}$

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Example Question #2 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ 4 x^{3} + 6 x^{2} + x + 2 }{ x + 8 }$

Possible Answers:

$4 x^{2} - 26 x + 209$

$- \frac{1670}{x + 8}$

$4 x^{2} - 26 x + 209 - \frac{1670}{x + 8}$

Correct answer:

$4 x^{2} - 26 x + 209 - \frac{1670}{x + 8}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 4 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & ~ & ~\\ \hline ~ & 4 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & ~ & ~\\ \hline ~ & 4 & -26 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & 208 & ~\\ \hline ~ & 4 & -26 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & 208 & ~\\ \hline ~ & 4 & -26 & 209 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & 208 & -1672 \\ \hline ~ & 4 & -26 & 209 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -8 & 4 & 6 & 1 & 2 \\ ~ & ~ & -32 & 208 & -1672 \\ \hline ~ & 4 & -26 & 209 & -1670 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = 4 x^{2} - 26 x + 209$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -1670$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x + 8$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = 4 x^{2} - 26 x + 209 - \frac{1670}{x + 8}$

Report an Error

Example Question #3 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ x^{3} - 4 x - 5 }{ x + 2 }$

Possible Answers:

$x^{2} - 2 x$

$x^{2} - 2 x - \frac{5}{x + 2}$

$- \frac{5}{x + 2}$

Correct answer:

$x^{2} - 2 x - \frac{5}{x + 2}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & ~ & ~\\ \hline ~ & 1 & -2 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & 4 & ~\\ \hline ~ & 1 & -2 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & 4 & ~\\ \hline ~ & 1 & -2 & 0 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & 4 & 0 \\ \hline ~ & 1 & -2 & 0 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -2 & 1 & 0 & -4 & -5 \\ ~ & ~ & -2 & 4 & 0 \\ \hline ~ & 1 & -2 & 0 & -5 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = x^{2} - 2 x$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -5$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x + 2$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = x^{2} - 2 x - \frac{5}{x + 2}$

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Example Question #383 : High School: Algebra

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ x^{3} + 7 x + 5 }{ x + 7 }$

Possible Answers:

$- \frac{387}{x + 7}$

$x^{2} - 7 x + 56 - \frac{387}{x + 7}$

$x^{2} - 7 x + 56$

Correct answer:

$x^{2} - 7 x + 56 - \frac{387}{x + 7}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & ~ & ~\\ \hline ~ & 1 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & ~ & ~\\ \hline ~ & 1 & -7 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & 49 & ~\\ \hline ~ & 1 & -7 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & 49 & ~\\ \hline ~ & 1 & -7 & 56 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & 49 & -392 \\ \hline ~ & 1 & -7 & 56 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -7 & 1 & 0 & 7 & 5 \\ ~ & ~ & -7 & 49 & -392 \\ \hline ~ & 1 & -7 & 56 & -387 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = x^{2} - 7 x + 56$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -387$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x + 7$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = x^{2} - 7 x + 56 - \frac{387}{x + 7}$

Report an Error

Example Question #6 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ 7 x^{3} - x^{2} + 5 x - 8 }{ x }$

Possible Answers:

$- \frac{8}{x}$

$7 x^{2} - x + 5$

$7 x^{2} - x + 5 - \frac{8}{x}$

Correct answer:

$7 x^{2} - x + 5 - \frac{8}{x}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & ~ & ~\\ \hline ~ & 7 & -1 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & 0 & ~\\ \hline ~ & 7 & -1 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & 0 & ~\\ \hline ~ & 7 & -1 & 5 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & 0 & 0 \\ \hline ~ & 7 & -1 & 5 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 0 & 7 & -1 & 5 & -8 \\ ~ & ~ & 0 & 0 & 0 \\ \hline ~ & 7 & -1 & 5 & -8 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = 7 x^{2} - x + 5$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -8$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = 7 x^{2} - x + 5 - \frac{8}{x}$

Report an Error

Example Question #4 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ - 3 x^{3} - 2 x^{2} - 5 x - 2 }{ x + 6 }$

Possible Answers:

$- 3 x^{2} + 16 x - 101 + \frac{604}{x + 6}$

516

$\frac{604}{x + 6}$

$- 3 x^{2} + 16 x - 101$

604

Correct answer:

$- 3 x^{2} + 16 x - 101 + \frac{604}{x + 6}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & -3 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & ~ & ~\\ \hline ~ & -3 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & ~ & ~\\ \hline ~ & -3 & 16 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & -96 & ~\\ \hline ~ & -3 & 16 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & -96 & ~\\ \hline ~ & -3 & 16 & -101 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & -96 & 606 \\ \hline ~ & -3 & 16 & -101 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} -6 & -3 & -2 & -5 & -2 \\ ~ & ~ & 18 & -96 & 606 \\ \hline ~ & -3 & 16 & -101 & 604 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = - 3 x^{2} + 16 x - 101$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = 604$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x + 6$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = - 3 x^{2} + 16 x - 101 + \frac{604}{x + 6}$

Report an Error

Example Question #5 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ - 4 x^{3} + 5 x^{2} - 4 x - 8 }{ x - 9 }$

Possible Answers:

$- 4 x^{2} - 31 x - 283 - \frac{2555}{x - 9}$

$- 4 x^{2} - 31 x - 283$

$- \frac{2555}{x - 9}$

Correct answer:

$- 4 x^{2} - 31 x - 283 - \frac{2555}{x - 9}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & -4 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & ~ & ~\\ \hline ~ & -4 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & ~ & ~\\ \hline ~ & -4 & -31 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & -279 & ~\\ \hline ~ & -4 & -31 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & -279 & ~\\ \hline ~ & -4 & -31 & -283 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & -279 & -2547 \\ \hline ~ & -4 & -31 & -283 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 9 & -4 & 5 & -4 & -8 \\ ~ & ~ & -36 & -279 & -2547 \\ \hline ~ & -4 & -31 & -283 & -2555 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = - 4 x^{2} - 31 x - 283$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -2555$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x - 9$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = - 4 x^{2} - 31 x - 283 - \frac{2555}{x - 9}$

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Example Question #3 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ 7 x^{3} + 3 x^{2} - x + 2 }{ x - 3 }$

Possible Answers:

215

$7 x^{2} + 24 x + 71$

$7 x^{2} + 24 x + 71 + \frac{215}{x - 3}$

317

$\frac{215}{x - 3}$

Correct answer:

$7 x^{2} + 24 x + 71 + \frac{215}{x - 3}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.
$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & ~ & ~\\ \hline ~ & 7 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & ~ & ~\\ \hline ~ & 7 & 24 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & 72 & ~\\ \hline ~ & 7 & 24 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & 72 & ~\\ \hline ~ & 7 & 24 & 71 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & 72 & 213 \\ \hline ~ & 7 & 24 & 71 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 3 & 7 & 3 & -1 & 2 \\ ~ & ~ & 21 & 72 & 213 \\ \hline ~ & 7 & 24 & 71 & 215 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = 7 x^{2} + 24 x + 71$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = 215$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x - 3$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = 7 x^{2} + 24 x + 71 + \frac{215}{x - 3}$

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Example Question #4 : Different Forms Of Simple Rational Expressions: Ccss.Math.Content.Hsa Apr.D.6

Write the following polynomial quotient in the form $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$\frac{ - 6 x^{3} + 2 x^{2} + 2 x + 1 }{ x - 1 }$

Possible Answers:

$- 6 x^{2} - 4 x - 2 - \frac{1}{x - 1}$

$- \frac{1}{x - 1}$

$- 6 x^{2} - 4 x - 2$

Correct answer:

$- 6 x^{2} - 4 x - 2 - \frac{1}{x - 1}$

Explanation:

In order to solve this problem, we need to perform synthetic division.

We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^3$ term down.

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & ~ & ~ & ~\\ \hline ~ & -6 & ~ & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}^2$ term coefficient.

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & ~ & ~\\ \hline ~ & -6 & ~ & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & ~ & ~\\ \hline ~ & -6 & -4 & ~ & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the $\uptext{x}$ term.

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & -4 & ~\\ \hline ~ & -6 & -4 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & -4 & ~\\ \hline ~ & -6 & -4 & -2 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & -4 & -2 \\ \hline ~ & -6 & -4 & -2 & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| cccc} 1 & -6 & 2 & 2 & 1 \\ ~ & ~ & -6 & -4 & -2 \\ \hline ~ & -6 & -4 & -2 & -1 \end{tabular}$

Now we need to write it out in the form of $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

$q{\left (x \right )}$ is the quotient, which is the first numbers from the synthetic division.

$q{\left (x \right )} = - 6 x^{2} - 4 x - 2$

$r{\left (x \right )}$ is the remainder, which is the last number in the synthetic division.

$r{\left (x \right )} = -1$

$b{\left (x \right )}$ is the divisor, which is what we originally divided by.

$b{\left (x \right )} = x - 1$

Now we put this all together to get.

$q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}} = - 6 x^{2} - 4 x - 2 - \frac{1}{x - 1}$

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Common Core: High School - Algebra : Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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