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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

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

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #62 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 8 x^{2} + 12 x + 9$ is divided by x - 10

Possible Answers:

104

100

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.

$\begin{tabular}{ c| ccc}10&-8&12&9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&-8&12&9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&-8&12&9\\ ~ & ~ &-80& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}10&-8&12&9\\~ & ~ &-80& ~ \\ \hline ~ &-8&-68& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}10&-8&12&9\\ ~ & ~ &-80&-680\\ \hline ~ &-8&-68& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}10&-8&12&9\\ ~ & ~ &-80&-680\\ \hline ~ &-8&-68&-671\end{tabular}$

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

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Example Question #63 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $x^{2} - 12 x - 8$ is divided by x + 14

Possible Answers:

364

196

185

Correct answer:

364

Explanation:

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\ ~ & ~ & ~ & ~ \\ \hline ~ &1& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\ ~ & ~ &-14& ~ \\ \hline ~ &1& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\~ & ~ &-14& ~ \\ \hline ~ &1&-26& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\ ~ & ~ &-14&364\\ \hline ~ &1&-26& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-14&1&-12&-8\\ ~ & ~ &-14&364\\ \hline ~ &1&-26&356\end{tabular}$

The last number is the remainder, so our final answer is 356 .

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Example Question #71 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 19 x^{2} - x + 1$ is divided by x - 7

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}7&-19&-1&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&-19&-1&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &-19& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&-19&-1&1\\ ~ & ~ &-133& ~ \\ \hline ~ &-19& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}7&-19&-1&1\\~ & ~ &-133& ~ \\ \hline ~ &-19&-134& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}7&-19&-1&1\\ ~ & ~ &-133&-938\\ \hline ~ &-19&-134& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}7&-19&-1&1\\ ~ & ~ &-133&-938\\ \hline ~ &-19&-134&-937\end{tabular}$

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

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Example Question #72 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- x^{2} + 9 x + 7$ is divided by x + 7

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}-7&-1&9&7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-7&-1&9&7\\ ~ & ~ & ~ & ~ \\ \hline ~ &-1& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-7&-1&9&7\\ ~ & ~ &7& ~ \\ \hline ~ &-1& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-7&-1&9&7\\~ & ~ &7& ~ \\ \hline ~ &-1&16& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-7&-1&9&7\\ ~ & ~ &7&-112\\ \hline ~ &-1&16& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-7&-1&9&7\\ ~ & ~ &7&-112\\ \hline ~ &-1&16&-105\end{tabular}$

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

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Example Question #73 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 18 x^{2} + 10 x - 7$ is divided by x - 6

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}6&-18&10&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}6&-18&10&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}6&-18&10&-7\\ ~ & ~ &-108& ~ \\ \hline ~ &-18& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}6&-18&10&-7\\~ & ~ &-108& ~ \\ \hline ~ &-18&-98& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}6&-18&10&-7\\ ~ & ~ &-108&-588\\ \hline ~ &-18&-98& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}6&-18&10&-7\\ ~ & ~ &-108&-588\\ \hline ~ &-18&-98&-595\end{tabular}$

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

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Example Question #74 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $19 x^{2} - 20 x + 4$ is divided by x - 13

Possible Answers:

2951

169

168

Correct answer:

2951

Explanation:

$\begin{tabular}{ c| ccc}13&19&-20&4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&19&-20&4\\ ~ & ~ & ~ & ~ \\ \hline ~ &19& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&19&-20&4\\ ~ & ~ &247& ~ \\ \hline ~ &19& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}13&19&-20&4\\~ & ~ &247& ~ \\ \hline ~ &19&227& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}13&19&-20&4\\ ~ & ~ &247&2951\\ \hline ~ &19&227& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}13&19&-20&4\\ ~ & ~ &247&2951\\ \hline ~ &19&227&2955\end{tabular}$

The last number is the remainder, so our final answer is 2955 .

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Example Question #75 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 16 x^{2} + 3 x + 6$ is divided by x + 4

Possible Answers:

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}-4&-16&3&6\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-4&-16&3&6\\ ~ & ~ & ~ & ~ \\ \hline ~ &-16& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-4&-16&3&6\\ ~ & ~ &64& ~ \\ \hline ~ &-16& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-4&-16&3&6\\~ & ~ &64& ~ \\ \hline ~ &-16&67& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-4&-16&3&6\\ ~ & ~ &64&-268\\ \hline ~ &-16&67& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-4&-16&3&6\\ ~ & ~ &64&-268\\ \hline ~ &-16&67&-262\end{tabular}$

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

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Example Question #76 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 19 x^{2} - 3 x + 7$ is divided by x - 15

Possible Answers:

203

225

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}15&-19&-3&7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-19&-3&7\\ ~ & ~ & ~ & ~ \\ \hline ~ &-19& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-19&-3&7\\ ~ & ~ &-285& ~ \\ \hline ~ &-19& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}15&-19&-3&7\\~ & ~ &-285& ~ \\ \hline ~ &-19&-288& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-19&-3&7\\ ~ & ~ &-285&-4320\\ \hline ~ &-19&-288& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}15&-19&-3&7\\ ~ & ~ &-285&-4320\\ \hline ~ &-19&-288&-4313\end{tabular}$

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

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Example Question #77 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

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

Possible Answers:

147

276

144

Correct answer:

276

Explanation:

$\begin{tabular}{ c| ccc}-12&2&1&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&2&1&-5\\ ~ & ~ & ~ & ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&2&1&-5\\ ~ & ~ &-24& ~ \\ \hline ~ &2& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}-12&2&1&-5\\~ & ~ &-24& ~ \\ \hline ~ &2&-23& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}-12&2&1&-5\\ ~ & ~ &-24&276\\ \hline ~ &2&-23& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}-12&2&1&-5\\ ~ & ~ &-24&276\\ \hline ~ &2&-23&271\end{tabular}$

The last number is the remainder, so our final answer is 271 .

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Example Question #78 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

What is the remainder when $- 9 x^{2} - 6 x - 9$ is divided by x - 15

Possible Answers:

210

225

Correct answer:

Explanation:

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-9& ~ & ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\ ~ & ~ &-135& ~ \\ \hline ~ &-9& ~ & ~ \end{tabular}$

Now we add the column up to get.

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\~ & ~ &-135& ~ \\ \hline ~ &-9&-141& ~ \end{tabular}$

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

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\ ~ & ~ &-135&-2115\\ \hline ~ &-9&-141& ~ \end{tabular}$

Now we add the column together to get.

$\begin{tabular}{ c| ccc}15&-9&-6&-9\\ ~ & ~ &-135&-2115\\ \hline ~ &-9&-141&-2124\end{tabular}$

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

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Arithmetic with Polynomials & Rational Expressions

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #62 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #63 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #71 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #72 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #73 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #74 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #75 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #76 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #77 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

Example Question #78 : Remainder Theorem: Ccss.Math.Content.Hsa Apr.B.2

All Common Core: High School - Algebra Resources

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