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

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Example Questions

Example Question #17 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $13 x^{2} + 9 x - 6$ $\displaystyle 13 x^{2} + 9 x - 6$ is divided by x - 10 $\displaystyle x - 10$

Possible Answers:

1390 $\displaystyle 1390$

122 $\displaystyle 122$

$\displaystyle 22$

$\displaystyle 16$

100 $\displaystyle 100$

Correct answer:

1390 $\displaystyle 1390$

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.

$\begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ &13& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ & ~ & ~ \\ \hline ~ &13& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130& ~ \\ \hline ~ &13& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130& ~ \\ \hline ~ &13& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}10&13&9&-6\\~ & ~ &130& ~ \\ \hline ~ &13&139& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\~ & ~ &130& ~ \\ \hline ~ &13&139& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130&1390\\ \hline ~ &13&139& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130&1390\\ \hline ~ &13&139& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130&1390\\ \hline ~ &13&139&1384\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}10&13&9&-6\\ ~ & ~ &130&1390\\ \hline ~ &13&139&1384\end{tabular}$

The last number is the remainder, so our final answer is 1384 $\displaystyle 1384$.

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Example Question #18 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 11 x^{2} + 19 x + 4$ $\displaystyle - 11 x^{2} + 19 x + 4$ is divided by x + 6 $\displaystyle x + 6$

Possible Answers:

$\displaystyle -510$

$\displaystyle 36$

$\displaystyle 8$

$\displaystyle 12$

$\displaystyle 44$

Correct answer:

$\displaystyle -510$

Explanation:

$\begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66& ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66& ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-6&-11&19&4\\~ & ~ &66& ~ \\ \hline ~ &-11&85& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\~ & ~ &66& ~ \\ \hline ~ &-11&85& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66&-510\\ \hline ~ &-11&85& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66&-510\\ \hline ~ &-11&85& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66&-510\\ \hline ~ &-11&85&-506\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-6&-11&19&4\\ ~ & ~ &66&-510\\ \hline ~ &-11&85&-506\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -506$.

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Example Question #31 : Arithmetic With Polynomials & Rational Expressions

What is the remainder when $- 11 x^{2} + 19 x + 4$ $\displaystyle - 11 x^{2} + 19 x + 4$ is divided by x - 13 $\displaystyle x - 13$

Possible Answers:

169 $\displaystyle 169$

$\displaystyle -1612$

$\displaystyle 8$

177 $\displaystyle 177$

$\displaystyle 12$

Correct answer:

$\displaystyle -1612$

Explanation:

$\begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143& ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143& ~ \\ \hline ~ &-11& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}13&-11&19&4\\~ & ~ &-143& ~ \\ \hline ~ &-11&-124& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\~ & ~ &-143& ~ \\ \hline ~ &-11&-124& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143&-1612\\ \hline ~ &-11&-124& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143&-1612\\ \hline ~ &-11&-124& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143&-1612\\ \hline ~ &-11&-124&-1608\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}13&-11&19&4\\ ~ & ~ &-143&-1612\\ \hline ~ &-11&-124&-1608\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -1608$.

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Example Question #101 : High School: Algebra

What is the remainder when $2 x^{2} - 19 x + 8$ $\displaystyle 2 x^{2} - 19 x + 8$ is divided by x - 18 $\displaystyle x - 18$

Possible Answers:

307 $\displaystyle 307$

$\displaystyle -9$

$\displaystyle -17$

324 $\displaystyle 324$

306 $\displaystyle 306$

Correct answer:

306 $\displaystyle 306$

Explanation:

$\begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}18&2&-19&8\\~ & ~ &36& ~ \\ \hline ~ &2&17& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\~ & ~ &36& ~ \\ \hline ~ &2&17& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36&306\\ \hline ~ &2&17& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36&306\\ \hline ~ &2&17& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36&306\\ \hline ~ &2&17&314\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}18&2&-19&8\\ ~ & ~ &36&306\\ \hline ~ &2&17&314\end{tabular}$

The last number is the remainder, so our final answer is 314 $\displaystyle 314$.

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Example Question #21 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 20 x^{2} - 5 x - 9$ $\displaystyle - 20 x^{2} - 5 x - 9$ is divided by x + 8 $\displaystyle x + 8$

Possible Answers:

$\displaystyle -34$

$\displaystyle 39$

$\displaystyle 64$

$\displaystyle -1240$

$\displaystyle -25$

Correct answer:

$\displaystyle -1240$

Explanation:

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-20& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-20& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160& ~ \\ \hline ~ &-20& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160& ~ \\ \hline ~ &-20& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\~ & ~ &160& ~ \\ \hline ~ &-20&155& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\~ & ~ &160& ~ \\ \hline ~ &-20&155& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160&-1240\\ \hline ~ &-20&155& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160&-1240\\ \hline ~ &-20&155& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160&-1240\\ \hline ~ &-20&155&-1249\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}-8&-20&-5&-9\\ ~ & ~ &160&-1240\\ \hline ~ &-20&155&-1249\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -1249$.

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Example Question #22 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 2 x^{2} + 19 x + 4$ $\displaystyle - 2 x^{2} + 19 x + 4$ is divided by x - 8 $\displaystyle x - 8$

Possible Answers:

$\displaystyle 24$

$\displaystyle 81$

$\displaystyle 17$

$\displaystyle 64$

$\displaystyle 21$

Correct answer:

$\displaystyle 24$

Explanation:

$\begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16& ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16& ~ \\ \hline ~ &-2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}8&-2&19&4\\~ & ~ &-16& ~ \\ \hline ~ &-2&3& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\~ & ~ &-16& ~ \\ \hline ~ &-2&3& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16&24\\ \hline ~ &-2&3& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16&24\\ \hline ~ &-2&3& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16&24\\ \hline ~ &-2&3&28\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-2&19&4\\ ~ & ~ &-16&24\\ \hline ~ &-2&3&28\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle 28$.

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Example Question #23 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 15 x^{2} + 9 x - 3$ $\displaystyle - 15 x^{2} + 9 x - 3$ is divided by x - 3 $\displaystyle x - 3$

Possible Answers:

$\displaystyle -108$

$\displaystyle 3$

$\displaystyle 9$

$\displaystyle -9$

$\displaystyle -6$

Correct answer:

$\displaystyle -108$

Explanation:

$\begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-15& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ & ~ & ~ \\ \hline ~ &-15& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45& ~ \\ \hline ~ &-15& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45& ~ \\ \hline ~ &-15& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}3&-15&9&-3\\~ & ~ &-45& ~ \\ \hline ~ &-15&-36& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\~ & ~ &-45& ~ \\ \hline ~ &-15&-36& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45&-108\\ \hline ~ &-15&-36& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45&-108\\ \hline ~ &-15&-36& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45&-108\\ \hline ~ &-15&-36&-111\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}3&-15&9&-3\\ ~ & ~ &-45&-108\\ \hline ~ &-15&-36&-111\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -111$.

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Example Question #24 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

$\displaystyle -13$

$\displaystyle -15$

$\displaystyle 4$

$\displaystyle -18$

$\displaystyle -9$

Correct answer:

$\displaystyle -18$

Explanation:

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

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

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

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

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

Now we add the column up to get.

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

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

$\begin{tabular}{ c| ccc}2&4&-17&-2\\ ~ & ~ &8&-18\\ \hline ~ &4&-9& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}2&4&-17&-2\\ ~ & ~ &8&-18\\ \hline ~ &4&-9& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}2&4&-17&-2\\ ~ & ~ &8&-18\\ \hline ~ &4&-9&-20\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}2&4&-17&-2\\ ~ & ~ &8&-18\\ \hline ~ &4&-9&-20\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -20$.

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Example Question #25 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $7 x^{2} - 3 x + 1$ $\displaystyle 7 x^{2} - 3 x + 1$ is divided by x - 16 $\displaystyle x - 16$

Possible Answers:

$\displaystyle 5$

256 $\displaystyle 256$

260 $\displaystyle 260$

$\displaystyle 4$

1744 $\displaystyle 1744$

Correct answer:

1744 $\displaystyle 1744$

Explanation:

$\begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &7& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &7& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112& ~ \\ \hline ~ &7& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112& ~ \\ \hline ~ &7& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}16&7&-3&1\\~ & ~ &112& ~ \\ \hline ~ &7&109& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\~ & ~ &112& ~ \\ \hline ~ &7&109& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112&1744\\ \hline ~ &7&109& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112&1744\\ \hline ~ &7&109& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112&1744\\ \hline ~ &7&109&1745\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}16&7&-3&1\\ ~ & ~ &112&1744\\ \hline ~ &7&109&1745\end{tabular}$

The last number is the remainder, so our final answer is 1745 $\displaystyle 1745$.

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Example Question #26 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

$\displaystyle -464$

$\displaystyle 64$

$\displaystyle -10$

$\displaystyle -2$

$\displaystyle 62$

Correct answer:

$\displaystyle -464$

Explanation:

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

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

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

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

$\begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}8&-8&6&-8\\~ & ~ &-64& ~ \\ \hline ~ &-8&-58& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&6&-8\\~ & ~ &-64& ~ \\ \hline ~ &-8&-58& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64&-464\\ \hline ~ &-8&-58& ~ \end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64&-464\\ \hline ~ &-8&-58& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64&-464\\ \hline ~ &-8&-58&-472\end{tabular}$ $\displaystyle \begin{tabular}{ c| ccc}8&-8&6&-8\\ ~ & ~ &-64&-464\\ \hline ~ &-8&-58&-472\end{tabular}$

The last number is the remainder, so our final answer is $\displaystyle -472$.

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

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All Common Core: High School - Algebra Resources

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Example Question #17 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #18 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #31 : Arithmetic With Polynomials & Rational Expressions

Example Question #101 : High School: Algebra

Example Question #21 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #22 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #23 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #24 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #25 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #26 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

All Common Core: High School - Algebra Resources

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