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Common Core: High School - Algebra : Remainder Theorem: CCSS.Math.Content.HSA-APR.B.2

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Example Question #1 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $9 x^{2} + 2 x$ is divided by x - 1

Possible Answers:

Correct answer:

Explanation:

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

$\begin{tabular}{ c| ccc} 1 & 9 & 2 & 0 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 9 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 1 & 9 & 2 & 0 \\ ~ & ~ & 9 & ~ \\ \hline ~ & 9 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} 1 & 9 & 2 & 0 \\ ~ & ~ & 9 & ~ \\ \hline ~ & 9 & 11 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} 1 & 9 & 2 & 0 \\ ~ & ~ & 9 & 11 \\ \hline ~ & 9 & 11 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} 1 & 9 & 2 & 0 \\ ~ & ~ & 9 & 11 \\ \hline ~ & 9 & 11 & 11 \end{tabular}$

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

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

What is the remainder when x^2- 18 x + 3 is divided by x - 9

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.

$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$
The first step is to bring the coefficient of the $\uptext{x}^2$ term down.

$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 1 & ~ & ~ \end{tabular}$

Now we multiply the zero by the term we just put down, and place it under the $\uptext{x}$ term coefficient.
$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & 9 & ~ \\ \hline ~ & 1 & ~ & ~ \end{tabular}$
Now we add the column up to get

$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & 9 & ~ \\ \hline ~ & 1 & -9& ~ \end{tabular}$
Now we multiply the number we got by the zero, and place it under the constant term.

$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & 9 & -81 \\ \hline ~ & 1 & -9 & ~ \end{tabular}$
Now we add the column together to get.

$\begin{tabular}{ c| ccc} 9 & 1 & -18 & 3 \\ ~ & ~ & 9 & -81 \\ \hline ~ & 1 & -9 &-78 \end{tabular}$

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

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Example Question #3 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $19 x^{2} + 4 x - 1$ is divided by x + 17

Possible Answers:

289

5422

312

Correct answer:

5422

Explanation:

$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 19 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & -323 & ~ \\ \hline ~ & 19 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & -323 & ~ \\ \hline ~ & 19 & -319 & ~ \end{tabular}$

Now we multiply the number we got by the zero, and place it under the constant term.
$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & -323 & 5423 \\ \hline ~ & 19 & -319 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} -17 & 19 & 4 & -1 \\ ~ & ~ & -323 & 5423 \\ \hline ~ & 19 & -319 & 5422 \end{tabular}$

The last number is the remainder, so our final answer is 5422

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Example Question #4 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 15 x^{2} + 18 x + 3$ is divided by x + 1

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & 15 & -33 \\ \hline ~ & -15 & 33 & -30 \end{tabular}$ 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.

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & ~ & ~ \\ \hline ~ & -15 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & 15 & ~ \\ \hline ~ & -15 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & 15 & ~ \\ \hline ~ & -15 & 33 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & 15 & -33 \\ \hline ~ & -15 & 33 & ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc} -1 & -15 & 18 & 3 \\ ~ & ~ & 15 & -33 \\ \hline ~ & -15 & 33 & -30 \end{tabular}$

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

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Example Question #5 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $19 x^{2} + x + 6$ is divided by x + 18

Possible Answers:

324

344

6144

Correct answer:

6144

Explanation:

$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & ~ & ~ \\ \hline ~ & 19 & ~ & ~ \end{tabular}$
Now we multiply the zero by the term we just put down, and place it under the \uptext{x} term coefficient.

$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & -342 & ~ \\ \hline ~ & 19 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & -342 & ~ \\ \hline ~ & 19 & -341 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & -342 & 6138 \\ \hline ~ & 19 & -341 & ~ \end{tabular}$

Now we add the column together to get.
$\begin{tabular}{ c| ccc} -18 & 19 & 1 & 6 \\ ~ & ~ & -342 & 6138 \\ \hline ~ & 19 & -341 & 6144 \end{tabular}$
The last number is the remainder, so our final answer is 6144

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Example Question #6 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 12 x^{2} - 15 x + 5$ is divided by x + 17

Possible Answers:

289

262

Correct answer:

Explanation:

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

$\begin{tabular}{ c| ccc} -17 & -12 & -15 & 5 \\ ~ & ~ & ~ & ~ \\ \hline ~ & -12 & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -17 & -12 & -15 & 5 \\ ~ & ~ & 204 & ~ \\ \hline ~ & -12 & ~ & ~ \end{tabular}$

Now we add the column up to get

$\begin{tabular}{ c| ccc} -17 & -12 & -15 & 5 \\ ~ & ~ & 204 & ~ \\ \hline ~ & -12 & 189 & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc} -17 & -12 & -15 & 5 \\ ~ & ~ & 204 & -3213 \\ \hline ~ & -12 & 189 & ~ \end{tabular}$
Now we add the column together to get.

$\begin{tabular}{ c| ccc} -17 & -12 & -15 & 5 \\ ~ & ~ & 204 & -3213 \\ \hline ~ & -12 & 189 & -3208 \end{tabular}$

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

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Example Question #7 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

4269

225

228

Correct answer:

4269

Explanation:

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

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 #9 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

169

184

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 #10 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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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Common Core: High School - Algebra : Remainder Theorem: CCSS.Math.Content.HSA-APR.B.2

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #1 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Example Question #3 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #4 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #5 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #6 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #7 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Example Question #9 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #10 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

All Common Core: High School - Algebra Resources

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