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Common Core: High School - Algebra : High School: Algebra

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

What is the coefficient of $x^{11} y^{8}$ in the expansion of $\left(x + y\right)^{19}$ ?

Possible Answers:

75582

There is no coefficient

Correct answer:

75582

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 = 19 and i = 8

Now we compute the following

${\binom{n}{i}} = \binom{ 19 }{ 8 } = 75582$

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Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{14}$

Possible Answers:

$\\x^{14} + 14 x^{13} y + 91 x^{12} y^{2} + 364 x^{11} y^{3} + 1001 x^{10} y^{4} + 2002 x^{9} y^{5} + 3003 x^{8} y^{6} + 3432 x^{7} y^{7} + 3003 x^{6} y^{8} + 2002 x^{5} y^{9} + 1001 x^{4} y^{10} + 364 x^{3} y^{11} + 91 x^{2} y^{12} + 14 x y^{13} + y^{14}$

Not Possible

$x^{14} + 14 x y^{13} + y^{13}$

$x^{14} + 14 x^{13} y + 91 x^{12} y^{2}$

$x^{14} + y^{14}$

Correct answer:

Explanation:

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

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

Since the exponent in the question is 14 we can replace $\uptext{n}$ , with 14 .

Now our equation looks like

$\sum_{i=0}^{14} x^{- i + 14} y^{i} {\binom{14}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 14 }{ 0 } x ^{\left( 14 - 0 \right)} y ^{ 0 }= x^{14}$

Term 2 : $\binom{ 14 }{ 1 } x ^{\left( 14 - 1 \right)} y ^{ 1 }= 14 x^{13} y$

Term 3 : $\binom{ 14 }{ 2 } x ^{\left( 14 - 2 \right)} y ^{ 2 }= 91 x^{12} y^{2}$

Term 4 : $\binom{ 14 }{ 3 } x ^{\left( 14 - 3 \right)} y ^{ 3 }= 364 x^{11} y^{3}$

Term 5 : $\binom{ 14 }{ 4 } x ^{\left( 14 - 4 \right)} y ^{ 4 }= 1001 x^{10} y^{4}$

Term 6 : $\binom{ 14 }{ 5 } x ^{\left( 14 - 5 \right)} y ^{ 5 }= 2002 x^{9} y^{5}$

Term 7 : $\binom{ 14 }{ 6 } x ^{\left( 14 - 6 \right)} y ^{ 6 }= 3003 x^{8} y^{6}$

Term 8 : $\binom{ 14 }{ 7 } x ^{\left( 14 - 7 \right)} y ^{ 7 }= 3432 x^{7} y^{7}$

Term 9 : $\binom{ 14 }{ 8 } x ^{\left( 14 - 8 \right)} y ^{ 8 }= 3003 x^{6} y^{8}$

Term 10 : $\binom{ 14 }{ 9 } x ^{\left( 14 - 9 \right)} y ^{ 9 }= 2002 x^{5} y^{9}$

Term 11 : $\binom{ 14 }{ 10 } x ^{\left( 14 - 10 \right)} y ^{ 10 }= 1001 x^{4} y^{10}$

Term 12 : $\binom{ 14 }{ 11 } x ^{\left( 14 - 11 \right)} y ^{ 11 }= 364 x^{3} y^{11}$

Term 13 : $\binom{ 14 }{ 12 } x ^{\left( 14 - 12 \right)} y ^{ 12 }= 91 x^{2} y^{12}$

Term 14 : $\binom{ 14 }{ 13 } x ^{\left( 14 - 13 \right)} y ^{ 13 }= 14 x y^{13}$

Term 15 : $\binom{ 14 }{ 14 } x ^{\left( 14 - 14 \right)} y ^{ 14 }= y^{14}$

Now we combine the expressions and we get

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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 #381 : High School: Algebra

Use Pascal's Triangle to Expand

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

Possible Answers:

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

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

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

Not Possible

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 #382 : High School: Algebra

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

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

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

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

Possible Answers:

$\frac{6}{x}$

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

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

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.

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

Report an Error

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:

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

525

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

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

396

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 #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.

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

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Common Core: High School - Algebra : High School: Algebra

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

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

Example Question #381 : High School: Algebra

Example Question #382 : High School: Algebra

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 #383 : High School: Algebra

All Common Core: High School - Algebra Resources

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