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

What is the remainder when $16 x^{2} - 19 x - 7$ $\displaystyle 16 x^{2} - 19 x - 7$ is divided by x + 12 $\displaystyle x + 12$

Possible Answers:

141 $\displaystyle 141$

$\displaystyle -3$

144 $\displaystyle 144$

$\displaystyle -10$

2532 $\displaystyle 2532$

Correct answer:

2532 $\displaystyle 2532$

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.

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192& ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192& ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\~ & ~ &-192& ~ \\ \hline ~ &16&-211& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\~ & ~ &-192& ~ \\ \hline ~ &16&-211& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192&2532\\ \hline ~ &16&-211& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192&2532\\ \hline ~ &16&-211& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192&2532\\ \hline ~ &16&-211&2525\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-12&16&-19&-7\\ ~ & ~ &-192&2532\\ \hline ~ &16&-211&2525\end{tabular}$

The last number is the remainder, so our final answer is 2525 $\displaystyle 2525$.

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Example Question #42 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $8 x^{2} - 14 x + 9$ $\displaystyle 8 x^{2} - 14 x + 9$ is divided by x + 6 $\displaystyle x + 6$

Possible Answers:

$\displaystyle 30$

372 $\displaystyle 372$

$\displaystyle 36$

$\displaystyle -6$

$\displaystyle 3$

Correct answer:

372 $\displaystyle 372$

Explanation:

$\begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ & ~ & ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ & ~ & ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48& ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48& ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-6&8&-14&9\\~ & ~ &-48& ~ \\ \hline ~ &8&-62& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\~ & ~ &-48& ~ \\ \hline ~ &8&-62& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48&372\\ \hline ~ &8&-62& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48&372\\ \hline ~ &8&-62& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48&372\\ \hline ~ &8&-62&381\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&8&-14&9\\ ~ & ~ &-48&372\\ \hline ~ &8&-62&381\end{tabular}$

The last number is the remainder, so our final answer is 381 $\displaystyle 381$.

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Example Question #43 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 2 x^{2} + 5 x + 8$ $\displaystyle - 2 x^{2} + 5 x + 8$ is divided by x - 20 $\displaystyle x - 20$

Possible Answers:

403 $\displaystyle 403$

400 $\displaystyle 400$

$\displaystyle -700$

$\displaystyle 11$

$\displaystyle 3$

Correct answer:

$\displaystyle -700$

Explanation:

$\begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ & ~ & ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ & ~ & ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40& ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40& ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}20&-2&5&8\\~ & ~ &-40& ~ \\ \hline ~ &-2&-35& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\~ & ~ &-40& ~ \\ \hline ~ &-2&-35& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40&-700\\ \hline ~ &-2&-35& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40&-700\\ \hline ~ &-2&-35& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40&-700\\ \hline ~ &-2&-35&-692\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&-2&5&8\\ ~ & ~ &-40&-700\\ \hline ~ &-2&-35&-692\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -692$.

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Example Question #44 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 8 x^{2} + 10 x - 10$ $\displaystyle - 8 x^{2} + 10 x - 10$ is divided by x - 8 $\displaystyle x - 8$

Possible Answers:

$\displaystyle 66$

$\displaystyle -8$

$\displaystyle 64$

$\displaystyle -432$

$\displaystyle 2$

Correct answer:

$\displaystyle -432$

Explanation:

$\begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}8&-8&10&-10\\~ & ~ &-64& ~ \\ \hline ~ &-8&-54& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\~ & ~ &-64& ~ \\ \hline ~ &-8&-54& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64&-432\\ \hline ~ &-8&-54& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64&-432\\ \hline ~ &-8&-54& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64&-432\\ \hline ~ &-8&-54&-442\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&10&-10\\ ~ & ~ &-64&-432\\ \hline ~ &-8&-54&-442\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -442$.

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Example Question #45 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $5 x^{2} + 14 x + 1$ $\displaystyle 5 x^{2} + 14 x + 1$ is divided by x - 20 $\displaystyle x - 20$

Possible Answers:

400 $\displaystyle 400$

2280 $\displaystyle 2280$

$\displaystyle 20$

419 $\displaystyle 419$

$\displaystyle 19$

Correct answer:

2280 $\displaystyle 2280$

Explanation:

$\begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &5& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &5& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100& ~ \\ \hline ~ &5& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100& ~ \\ \hline ~ &5& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}20&5&14&1\\~ & ~ &100& ~ \\ \hline ~ &5&114& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\~ & ~ &100& ~ \\ \hline ~ &5&114& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100&2280\\ \hline ~ &5&114& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100&2280\\ \hline ~ &5&114& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100&2280\\ \hline ~ &5&114&2281\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}20&5&14&1\\ ~ & ~ &100&2280\\ \hline ~ &5&114&2281\end{tabular}$

The last number is the remainder, so our final answer is 2281 $\displaystyle 2281$.

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Example Question #46 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $14 x^{2} - 5 x + 6$ $\displaystyle 14 x^{2} - 5 x + 6$ is divided by x - 7 $\displaystyle x - 7$

Possible Answers:

$\displaystyle 15$

$\displaystyle 49$

$\displaystyle 9$

651 $\displaystyle 651$

$\displaystyle 58$

Correct answer:

651 $\displaystyle 651$

Explanation:

$\begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ & ~ & ~ \\ \hline ~ &14& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ & ~ & ~ \\ \hline ~ &14& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98& ~ \\ \hline ~ &14& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98& ~ \\ \hline ~ &14& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}7&14&-5&6\\~ & ~ &98& ~ \\ \hline ~ &14&93& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\~ & ~ &98& ~ \\ \hline ~ &14&93& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98&651\\ \hline ~ &14&93& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98&651\\ \hline ~ &14&93& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98&651\\ \hline ~ &14&93&657\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&14&-5&6\\ ~ & ~ &98&651\\ \hline ~ &14&93&657\end{tabular}$

The last number is the remainder, so our final answer is 657 $\displaystyle 657$.

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Example Question #51 : Arithmetic With Polynomials & Rational Expressions

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

Possible Answers:

225 $\displaystyle 225$

$\displaystyle -20$

$\displaystyle -3900$

203 $\displaystyle 203$

$\displaystyle -22$

Correct answer:

$\displaystyle -3900$

Explanation:

$\begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ & ~ & ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ & ~ & ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255& ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255& ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}15&-17&-5&2\\~ & ~ &-255& ~ \\ \hline ~ &-17&-260& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\~ & ~ &-255& ~ \\ \hline ~ &-17&-260& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255&-3900\\ \hline ~ &-17&-260& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255&-3900\\ \hline ~ &-17&-260& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255&-3900\\ \hline ~ &-17&-260&-3898\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&-17&-5&2\\ ~ & ~ &-255&-3900\\ \hline ~ &-17&-260&-3898\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -3898$.

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Example Question #52 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $- 13 x^{2} + 11 x - 10$ $\displaystyle - 13 x^{2} + 11 x - 10$ is divided by x - 13 $\displaystyle x - 13$

Possible Answers:

$\displaystyle -12$

$\displaystyle -2054$

$\displaystyle -2$

167 $\displaystyle 167$

169 $\displaystyle 169$

Correct answer:

$\displaystyle -2054$

Explanation:

$\begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ & ~ & ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169& ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169& ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}13&-13&11&-10\\~ & ~ &-169& ~ \\ \hline ~ &-13&-158& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\~ & ~ &-169& ~ \\ \hline ~ &-13&-158& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169&-2054\\ \hline ~ &-13&-158& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169&-2054\\ \hline ~ &-13&-158& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169&-2054\\ \hline ~ &-13&-158&-2064\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-13&11&-10\\ ~ & ~ &-169&-2054\\ \hline ~ &-13&-158&-2064\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -2064$.

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Example Question #61 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $8 x^{2} - x$ $\displaystyle 8 x^{2} - x$ is divided by x - 1 $\displaystyle x - 1$

Possible Answers:

$\displaystyle 7$

$\displaystyle 8$

$\displaystyle 1$

$\displaystyle 7$

Correct answer:

$\displaystyle 7$

Explanation:

$\begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ & ~ & ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ & ~ & ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8& ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8& ~ \\ \hline ~ &8& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}1&8&-1&0\\~ & ~ &8& ~ \\ \hline ~ &8&7& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\~ & ~ &8& ~ \\ \hline ~ &8&7& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8&7\\ \hline ~ &8&7& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8&7\\ \hline ~ &8&7& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8&7\\ \hline ~ &8&7&7\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}1&8&-1&0\\ ~ & ~ &8&7\\ \hline ~ &8&7&7\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle 7$.

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Example Question #62 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $15 x^{2} - 17 x + 3$ $\displaystyle 15 x^{2} - 17 x + 3$ is divided by x - 15 $\displaystyle x - 15$

Possible Answers:

225 $\displaystyle 225$

$\displaystyle -2$

223 $\displaystyle 223$

$\displaystyle 1$

3120 $\displaystyle 3120$

Correct answer:

3120 $\displaystyle 3120$

Explanation:

$\begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &15& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &15& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225& ~ \\ \hline ~ &15& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225& ~ \\ \hline ~ &15& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}15&15&-17&3\\~ & ~ &225& ~ \\ \hline ~ &15&208& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\~ & ~ &225& ~ \\ \hline ~ &15&208& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225&3120\\ \hline ~ &15&208& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225&3120\\ \hline ~ &15&208& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225&3120\\ \hline ~ &15&208&3123\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}15&15&-17&3\\ ~ & ~ &225&3120\\ \hline ~ &15&208&3123\end{tabular}$

The last number is the remainder, so our final answer is 3123 $\displaystyle 3123$.

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

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Example Question #41 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #42 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #43 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #44 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #45 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #46 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #51 : Arithmetic With Polynomials & Rational Expressions

Example Question #52 : Arithmetic With Polynomials & Rational Expressions

Example Question #61 : Arithmetic With Polynomials & Rational Expressions

Example Question #62 : Arithmetic With Polynomials & Rational Expressions

All Common Core: High School - Algebra Resources

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