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

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Example Questions

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

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

Possible Answers:

$\displaystyle 2$

$\displaystyle 12$

$\displaystyle 3$

$\displaystyle 1$

$\displaystyle 11$

Correct answer:

$\displaystyle 11$

Explanation:

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

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

The last number is the remainder, so our final answer is $\displaystyle 11$.

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

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

Possible Answers:

$\displaystyle -78$

$\displaystyle -87$

$\displaystyle -25$

$\displaystyle -16$

$\displaystyle -54$

Correct answer:

$\displaystyle -78$

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.

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

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

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

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

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

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

The last number is the remainder, so our final answer is $\displaystyle -78$

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

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

Possible Answers:

$\displaystyle 22$

5422 $\displaystyle 5422$

$\displaystyle 23$

312 $\displaystyle 312$

289 $\displaystyle 289$

Correct answer:

5422 $\displaystyle 5422$

Explanation:

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

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

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

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

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

Now we add the column up to get

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

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

Now we add the column together to get.

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

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

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

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

Possible Answers:

$\displaystyle 1$

$\displaystyle 4$

$\displaystyle 6$

$\displaystyle 3$

$\displaystyle -30$

Correct answer:

$\displaystyle -30$

Explanation:

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

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

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

The last number is the remainder, so our final answer is $\displaystyle -30$

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

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

Possible Answers:

$\displaystyle 20$

$\displaystyle 26$

324 $\displaystyle 324$

6144 $\displaystyle 6144$

344 $\displaystyle 344$

Correct answer:

6144 $\displaystyle 6144$

Explanation:

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

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

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

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

Now we add the column up to get

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

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

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

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

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

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

Possible Answers:

289 $\displaystyle 289$

262 $\displaystyle 262$

$\displaystyle -27$

$\displaystyle -3208$

$\displaystyle -22$

Correct answer:

$\displaystyle -3208$

Explanation:

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

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

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

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

Now we add the column up to get

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

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

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

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

The last number is the remainder, so our final answer is $\displaystyle -3208$

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

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

Possible Answers:

4269 $\displaystyle 4269$

$\displaystyle 3$

$\displaystyle -3$

228 $\displaystyle 228$

225 $\displaystyle 225$

Correct answer:

4269 $\displaystyle 4269$

Explanation:

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

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

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

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

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

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

Possible Answers:

121 $\displaystyle 121$

$\displaystyle -17$

$\displaystyle -625$

104 $\displaystyle 104$

$\displaystyle -15$

Correct answer:

$\displaystyle -625$

Explanation:

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

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

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

The last number is the remainder, so our final answer is $\displaystyle -625$.

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

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

Possible Answers:

$\displaystyle 15$

$\displaystyle -743$

$\displaystyle 13$

184 $\displaystyle 184$

169 $\displaystyle 169$

Correct answer:

$\displaystyle -743$

Explanation:

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

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

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

The last number is the remainder, so our final answer is $\displaystyle -743$

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

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

Possible Answers:

201 $\displaystyle 201$

$\displaystyle 3$

$\displaystyle 14$

$\displaystyle -2$

$\displaystyle 16$

Correct answer:

201 $\displaystyle 201$

Explanation:

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

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

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

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

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

Now we add the column up to get

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

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

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

Now we add the column together to get.

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

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

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