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All Common Core: High School - Algebra Resources

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Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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

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

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

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

$\frac{6}{x}$

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Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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

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

Your answer:

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

Your answer:

$\frac{6}{x}$

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.

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

Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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

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

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

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

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

396

525

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Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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

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

Your answer:

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

Your answer:

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

Your answer:

396

Your answer:

525

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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All Common Core: High School - Algebra Resources

Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

Flashcards: Different Forms of Simple Rational Expressions: CCSS.Math.Content.HSA-APR.D.6

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