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

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

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #91 : High School: Algebra

What is the remainder when $18 x^{2} - 15 x - 6$ is divided by x + 15

Possible Answers:

228

225

4269

Correct answer:

4269

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.

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 18 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & -270 & ~ \\ \hline ~ & 18 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & -270 & ~ \\ \hline ~ & 18 & -285 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & -270 & 4275 \\ \hline ~ & 18 & -285 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} -15 & 18 & -15 & -6 \\ ~ & ~ & -270 & 4275 \\ \hline ~ & 18 & -285 & 4269 \end{tabular}$

The last number is the remainder, so our final answer is 4269

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Example Question #8 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 4 x^{2} - 13 x + 2$ is divided by x - 11

Possible Answers:

121

104

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & ~ & ~ \\ \hline ~ & -4 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & -44 & ~ \\ \hline ~ & -4 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & -44 & ~ \\ \hline ~ & -4 & -57 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & -44 & -627 \\ \hline ~ & -4 & -57 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} 11 & -4 & -13 & 2 \\ ~ & ~ & -44 & -627 \\ \hline ~ & -4 & -57 & -625 \end{tabular}$

The last number is the remainder, so our final answer is .

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Example Question #21 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $- 3 x^{2} + 18 x - 2$ is divided by x + 13

Possible Answers:

184

169

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & ~ & ~ \\ \hline ~ & -3 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & 39 & ~ \\ \hline ~ & -3 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & 39 & ~ \\ \hline ~ & -3 & 57 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & 39 & -741 \\ \hline ~ & -3 & 57 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} -13 & -3 & 18 & -2 \\ ~ & ~ & 39 & -741 \\ \hline ~ & -3 & 57 & -743 \end{tabular}$

The last number is the remainder, so our final answer is

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Example Question #22 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $17 x^{2} - 19 x + 5$ is divided by x - 4

Possible Answers:

201

Correct answer:

201

Explanation:

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 17 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & 68 & ~ \\ \hline ~ & 17 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & 68 & ~ \\ \hline ~ & 17 & 49 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & 68 & 196 \\ \hline ~ & 17 & 49 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} 4 & 17 & -19 & 5 \\ ~ & ~ & 68 & 196 \\ \hline ~ & 17 & 49 & 201 \end{tabular}$

The last number is the remainder, so our final answer is 201

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Example Question #11 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 15 x^{2} - 8 x + 3$ is divided by x + 1

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc} -1 & -15 & -8 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} -1 & -15 & -8 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & -15 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -1 & -15 & -8 & 3 \\ ~ & ~ & 15 & ~ \\ \hline ~ & -15 & ~ & ~ \end{tabular}$

Now we add the column up to get

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

$\begin{tabular}{ c| ccc} -1 & -15 & -8 & 3 \\ ~ & ~ & 15 & -7 \\ \hline ~ & -15 & 7 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} -1 & -15 & -8 & 3 \\ ~ & ~ & 15 & -7 \\ \hline ~ & -15 & 7 & -4 \end{tabular}$

The last number is the remainder, so our final answer is .

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Example Question #12 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $12 x^{2} - 16 x - 5$ is divided by x - 13

Possible Answers:

169

1815

165

Correct answer:

1815

Explanation:

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 12 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & 156 & ~ \\ \hline ~ & 12 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & 156 & ~ \\ \hline ~ & 12 & 140 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & 156 & 1820 \\ \hline ~ & 12 & 140 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} 13 & 12 & -16 & -5 \\ ~ & ~ & 156 & 1820 \\ \hline ~ & 12 & 140 & 1815 \end{tabular}$

The last number is the remainder, so our final answer is 1815 .

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Example Question #13 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 14 x^{2} - 3 x + 2$ is divided by x + 17

Possible Answers:

289

272

Correct answer:

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.

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\ ~ & ~ & ~ & ~ \\ \hline ~ &-14& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\ ~ & ~ &238& ~ \\ \hline ~ &-14& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\~ & ~ &238& ~ \\ \hline ~ &-14&235& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\ ~ & ~ &238&-3995\\ \hline ~ &-14&235& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-17&-14&-3&2\\ ~ & ~ &238&-3995\\ \hline ~ &-14&235&-3993\end{tabular}$

The last number is the remainder, so our final answer is .

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Example Question #14 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 4 x^{2} - 4 x + 9$ is divided by x + 14

Possible Answers:

196

188

Correct answer:

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.

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-4& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\ ~ & ~ &56& ~ \\ \hline ~ &-4& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\~ & ~ &56& ~ \\ \hline ~ &-4&52& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\ ~ & ~ &56&-728\\ \hline ~ &-4&52& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-14&-4&-4&9\\ ~ & ~ &56&-728\\ \hline ~ &-4&52&-719\end{tabular}$

The last number is the remainder, so our final answer is .

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Example Question #15 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 11 x^{2} - 7 x - 8$ is divided by x + 13

Possible Answers:

151

169

Correct answer:

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.

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\ ~ & ~ & ~ & ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\ ~ & ~ &143& ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\~ & ~ &143& ~ \\ \hline ~ &-11&136& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\ ~ & ~ &143&-1768\\ \hline ~ &-11&136& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-13&-11&-7&-8\\ ~ & ~ &143&-1768\\ \hline ~ &-11&136&-1776\end{tabular}$

The last number is the remainder, so our final answer is .

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Example Question #16 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 2 x^{2} - 16 x + 6$ is divided by x + 7

Possible Answers:

Correct answer:

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.

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\ ~ & ~ & ~ & ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\ ~ & ~ &14& ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\~ & ~ &14& ~ \\ \hline ~ &-2&-2& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\ ~ & ~ &14&14\\ \hline ~ &-2&-2& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-7&-2&-16&6\\ ~ & ~ &14&14\\ \hline ~ &-2&-2&20\end{tabular}$

The last number is the remainder, so our final answer is .

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All Common Core: High School - Algebra Resources

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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 #91 : High School: Algebra

Example Question #8 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #21 : Arithmetic With Polynomials & Rational Expressions

Example Question #22 : Arithmetic With Polynomials & Rational Expressions

Example Question #11 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #12 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #13 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #14 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #15 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #16 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

All Common Core: High School - Algebra Resources

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