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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

Study concepts, example questions & explanations for Common Core: High School - Algebra

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Example Questions

Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{10} y^{4}$ in the expansion of $\left(x + y\right)^{14}$ ?

Possible Answers:

1001

There is no coefficient

Correct answer:

1001

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 14 and i = 4

Now we compute the following

${\binom{n}{i}} = \binom{ 14 }{ 4 } = 1001$

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Example Question #11 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{2}$

Possible Answers:

Not Possible

$x^{2} + y^{2}$

$x^{2} + 2 x y + y^{2}$

$x^{2} + 2 x y + y$

Correct answer:

$x^{2} + 2 x y + y^{2}$

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 2 we can replace $\uptext{n}$ , with 2 .

Now our equation looks like

$\sum_{i=0}^{2} x^{- i + 2} y^{i} {\binom{2}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 2 }{ 0 } x ^{\left( 2 - 0 \right)} y ^{ 0 }= x^{2}$

Term 2 : $\binom{ 2 }{ 1 } x ^{\left( 2 - 1 \right)} y ^{ 1 }= 2 x y$

Term 3 : $\binom{ 2 }{ 2 } x ^{\left( 2 - 2 \right)} y ^{ 2 }= y^{2}$

Now we combine the expressions and we get

$x^{2} + 2 x y + y^{2}$

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Example Question #12 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{12} y^{2}$ in the expansion of $\left(x + y\right)^{14}$ ?

Possible Answers:

There is no coefficient

Correct answer:

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 14 and i = 2

Now we compute the following

${\binom{n}{i}} = \binom{ 14 }{ 2 } = 91$

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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:

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

$\frac{6}{x}$

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

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.

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

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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 $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

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

Possible Answers:

525

396

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

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

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

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.

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

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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 $q{\left (x \right )} + \frac{r{\left (x \right )}}{b{\left (x \right )}}$

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

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

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

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

Possible Answers:

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

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

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

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.

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

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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.

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

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Example Question #11 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Example Question #12 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

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

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

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

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

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

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

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

All Common Core: High School - Algebra Resources

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