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

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Example Questions

Example Question #171 : High School: Algebra

What is the remainder when $- 18 x^{2} - 2 x - 3$ $\displaystyle - 18 x^{2} - 2 x - 3$ is divided by x - 5 $\displaystyle x - 5$

Possible Answers:

$\displaystyle 25$

$\displaystyle -20$

$\displaystyle -460$

$\displaystyle 5$

$\displaystyle -23$

Correct answer:

$\displaystyle -460$

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.

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90& ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90& ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\~ & ~ &-90& ~ \\ \hline ~ &-18&-92& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\~ & ~ &-90& ~ \\ \hline ~ &-18&-92& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90&-460\\ \hline ~ &-18&-92& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90&-460\\ \hline ~ &-18&-92& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90&-460\\ \hline ~ &-18&-92&-463\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}5&-18&-2&-3\\ ~ & ~ &-90&-460\\ \hline ~ &-18&-92&-463\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -463$.

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Example Question #172 : High School: Algebra

What is the remainder when $4 x^{2} - 9 x + 3$ $\displaystyle 4 x^{2} - 9 x + 3$ is divided by x + 2 $\displaystyle x + 2$

Possible Answers:

$\displaystyle 34$

$\displaystyle -1$

$\displaystyle -2$

$\displaystyle -5$

$\displaystyle 4$

Correct answer:

$\displaystyle 34$

Explanation:

$\begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &4& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &4& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8& ~ \\ \hline ~ &4& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8& ~ \\ \hline ~ &4& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-2&4&-9&3\\~ & ~ &-8& ~ \\ \hline ~ &4&-17& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\~ & ~ &-8& ~ \\ \hline ~ &4&-17& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8&34\\ \hline ~ &4&-17& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8&34\\ \hline ~ &4&-17& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8&34\\ \hline ~ &4&-17&37\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-2&4&-9&3\\ ~ & ~ &-8&34\\ \hline ~ &4&-17&37\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle 37$.

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Example Question #81 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

$\displaystyle 74$

$\displaystyle -7$

$\displaystyle 81$

$\displaystyle -13$

243 $\displaystyle 243$

Correct answer:

243 $\displaystyle 243$

Explanation:

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\~ & ~ &-18& ~ \\ \hline ~ &2&-27& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\~ & ~ &-18& ~ \\ \hline ~ &2&-27& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18&243\\ \hline ~ &2&-27& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18&243\\ \hline ~ &2&-27& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18&243\\ \hline ~ &2&-27&237\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-9&2&-9&-6\\ ~ & ~ &-18&243\\ \hline ~ &2&-27&237\end{tabular}$

The last number is the remainder, so our final answer is 237 $\displaystyle 237$.

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Example Question #174 : High School: Algebra

What is the remainder when $16 x^{2} - 14 x - 9$ $\displaystyle 16 x^{2} - 14 x - 9$ is divided by x - 12 $\displaystyle x - 12$

Possible Answers:

2136 $\displaystyle 2136$

$\displaystyle -7$

146 $\displaystyle 146$

$\displaystyle 2$

144 $\displaystyle 144$

Correct answer:

2136 $\displaystyle 2136$

Explanation:

$\begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192& ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192& ~ \\ \hline ~ &16& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}12&16&-14&-9\\~ & ~ &192& ~ \\ \hline ~ &16&178& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\~ & ~ &192& ~ \\ \hline ~ &16&178& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192&2136\\ \hline ~ &16&178& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192&2136\\ \hline ~ &16&178& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192&2136\\ \hline ~ &16&178&2127\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}12&16&-14&-9\\ ~ & ~ &192&2136\\ \hline ~ &16&178&2127\end{tabular}$

The last number is the remainder, so our final answer is 2127 $\displaystyle 2127$.

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Example Question #91 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $2 x^{2} + 13 x + 1$ $\displaystyle 2 x^{2} + 13 x + 1$ is divided by x - 7 $\displaystyle x - 7$

Possible Answers:

$\displaystyle 64$

$\displaystyle 15$

$\displaystyle 16$

189 $\displaystyle 189$

$\displaystyle 49$

Correct answer:

189 $\displaystyle 189$

Explanation:

$\begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}7&2&13&1\\~ & ~ &14& ~ \\ \hline ~ &2&27& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\~ & ~ &14& ~ \\ \hline ~ &2&27& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14&189\\ \hline ~ &2&27& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14&189\\ \hline ~ &2&27& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14&189\\ \hline ~ &2&27&190\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}7&2&13&1\\ ~ & ~ &14&189\\ \hline ~ &2&27&190\end{tabular}$

The last number is the remainder, so our final answer is 190 $\displaystyle 190$.

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Example Question #92 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $17 x^{2} - 20 x + 3$ $\displaystyle 17 x^{2} - 20 x + 3$ is divided by x - 16 $\displaystyle x - 16$

Possible Answers:

4032 $\displaystyle 4032$

256 $\displaystyle 256$

253 $\displaystyle 253$

$\displaystyle -3$

Correct answer:

4032 $\displaystyle 4032$

Explanation:

$\begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &17& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &17& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272& ~ \\ \hline ~ &17& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272& ~ \\ \hline ~ &17& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}16&17&-20&3\\~ & ~ &272& ~ \\ \hline ~ &17&252& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\~ & ~ &272& ~ \\ \hline ~ &17&252& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272&4032\\ \hline ~ &17&252& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272&4032\\ \hline ~ &17&252& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272&4032\\ \hline ~ &17&252&4035\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&17&-20&3\\ ~ & ~ &272&4032\\ \hline ~ &17&252&4035\end{tabular}$

The last number is the remainder, so our final answer is 4035 $\displaystyle 4035$.

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Example Question #93 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $12 x^{2} - 19 x - 7$ $\displaystyle 12 x^{2} - 19 x - 7$ is divided by x - 8 $\displaystyle x - 8$

Possible Answers:

$\displaystyle -7$

$\displaystyle 57$

616 $\displaystyle 616$

$\displaystyle -14$

$\displaystyle 64$

Correct answer:

616 $\displaystyle 616$

Explanation:

$\begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &12& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &12& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96& ~ \\ \hline ~ &12& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96& ~ \\ \hline ~ &12& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}8&12&-19&-7\\~ & ~ &96& ~ \\ \hline ~ &12&77& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\~ & ~ &96& ~ \\ \hline ~ &12&77& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96&616\\ \hline ~ &12&77& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96&616\\ \hline ~ &12&77& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96&616\\ \hline ~ &12&77&609\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&12&-19&-7\\ ~ & ~ &96&616\\ \hline ~ &12&77&609\end{tabular}$

The last number is the remainder, so our final answer is 609 $\displaystyle 609$.

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Example Question #94 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $18 x^{2} - 8 x - 7$ $\displaystyle 18 x^{2} - 8 x - 7$ is divided by x + 17 $\displaystyle x + 17$

Possible Answers:

5338 $\displaystyle 5338$

$\displaystyle 3$

$\displaystyle 10$

289 $\displaystyle 289$

299 $\displaystyle 299$

Correct answer:

5338 $\displaystyle 5338$

Explanation:

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &18& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &18& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306& ~ \\ \hline ~ &18& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306& ~ \\ \hline ~ &18& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\~ & ~ &-306& ~ \\ \hline ~ &18&-314& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\~ & ~ &-306& ~ \\ \hline ~ &18&-314& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306&5338\\ \hline ~ &18&-314& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306&5338\\ \hline ~ &18&-314& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306&5338\\ \hline ~ &18&-314&5331\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-17&18&-8&-7\\ ~ & ~ &-306&5338\\ \hline ~ &18&-314&5331\end{tabular}$

The last number is the remainder, so our final answer is 5331 $\displaystyle 5331$.

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Example Question #95 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 6 x^{2} - 4 x + 3$ $\displaystyle - 6 x^{2} - 4 x + 3$ is divided by x - 3 $\displaystyle x - 3$

Possible Answers:

$\displaystyle 9$

$\displaystyle -66$

$\displaystyle -10$

$\displaystyle -1$

$\displaystyle -7$

Correct answer:

$\displaystyle -66$

Explanation:

$\begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-6& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-6& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18& ~ \\ \hline ~ &-6& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18& ~ \\ \hline ~ &-6& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}3&-6&-4&3\\~ & ~ &-18& ~ \\ \hline ~ &-6&-22& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\~ & ~ &-18& ~ \\ \hline ~ &-6&-22& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18&-66\\ \hline ~ &-6&-22& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18&-66\\ \hline ~ &-6&-22& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18&-66\\ \hline ~ &-6&-22&-63\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-6&-4&3\\ ~ & ~ &-18&-66\\ \hline ~ &-6&-22&-63\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -63$.

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Example Question #96 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $6 x^{2} + 13 x - 5$ $\displaystyle 6 x^{2} + 13 x - 5$ is divided by x - 13 $\displaystyle x - 13$

Possible Answers:

$\displaystyle 19$

188 $\displaystyle 188$

1183 $\displaystyle 1183$

$\displaystyle 14$

169 $\displaystyle 169$

Correct answer:

1183 $\displaystyle 1183$

Explanation:

$\begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ &6& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ &6& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78& ~ \\ \hline ~ &6& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78& ~ \\ \hline ~ &6& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}13&6&13&-5\\~ & ~ &78& ~ \\ \hline ~ &6&91& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\~ & ~ &78& ~ \\ \hline ~ &6&91& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78&1183\\ \hline ~ &6&91& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78&1183\\ \hline ~ &6&91& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78&1183\\ \hline ~ &6&91&1178\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&6&13&-5\\ ~ & ~ &78&1183\\ \hline ~ &6&91&1178\end{tabular}$

The last number is the remainder, so our final answer is 1178 $\displaystyle 1178$.

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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 #171 : High School: Algebra

Example Question #172 : High School: Algebra

Example Question #81 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #174 : High School: Algebra

Example Question #91 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #92 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #93 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #94 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #95 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #96 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

All Common Core: High School - Algebra Resources

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