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Common Core: High School - Algebra : Polynomial Identities and Numerical Relationships: CCSS.Math.Content.HSA-APR.C.4

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Example Questions

Example Question #1 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 10 b\right)^{2}$ $\displaystyle \left(a + 10 b\right)^{2}$

Possible Answers:

$a^{2}$ $\displaystyle a^{2}$

$b^{2}$ $\displaystyle b^{2}$

$10 a b + 100 b^{2}$ $\displaystyle 10 a b + 100 b^{2}$

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

$a^{2} + 20 a b$ $\displaystyle a^{2} + 20 a b$

Correct answer:

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 10 b\right)^{2} = \left( a + 10*b \right) \cdot \left( a + 10*b \right)$ $\displaystyle \left(a + 10 b\right)^{2} = \left( a + 10*b \right) \cdot \left( a + 10*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 10 b = 10 a b$ $\displaystyle a \cdot 10 b = 10 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 10 b = 10 a b$ $\displaystyle a \cdot 10 b = 10 a b$

Now we multiply the last terms of each expression together.

$10 b \cdot 10 b = 100 b^{2}$ $\displaystyle 10 b \cdot 10 b = 100 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 20 a b + 100 b^{2}$ $\displaystyle a^{2} + 20 a b + 100 b^{2}$

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Example Question #2 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 14 b\right)^{2}$ $\displaystyle \left(a + 14 b\right)^{2}$

Possible Answers:

$a^{2}$ $\displaystyle a^{2}$

$b^{2}$ $\displaystyle b^{2}$

$14 a b + 196 b^{2}$ $\displaystyle 14 a b + 196 b^{2}$

$a^{2} + 28 a b + 196 b^{2}$ $\displaystyle a^{2} + 28 a b + 196 b^{2}$

$a^{2} + 28 a b$ $\displaystyle a^{2} + 28 a b$

Correct answer:

$a^{2} + 28 a b + 196 b^{2}$ $\displaystyle a^{2} + 28 a b + 196 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 14 b\right)^{2} = \left( a + 14*b \right) \cdot \left( a + 14*b \right)$ $\displaystyle \left(a + 14 b\right)^{2} = \left( a + 14*b \right) \cdot \left( a + 14*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 14 b = 14 a b$ $\displaystyle a \cdot 14 b = 14 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 14 b = 14 a b$ $\displaystyle a \cdot 14 b = 14 a b$

Now we multiply the last terms of each expression together.

$14 b \cdot 14 b = 196 b^{2}$ $\displaystyle 14 b \cdot 14 b = 196 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 28 a b + 196 b^{2}$ $\displaystyle a^{2} + 28 a b + 196 b^{2}$

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Example Question #3 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 2 b\right)^{2}$ $\displaystyle \left(a + 2 b\right)^{2}$

Possible Answers:

$a^{2} + 4 a b$ $\displaystyle a^{2} + 4 a b$

$b^{2}$ $\displaystyle b^{2}$

$2 a b + 4 b^{2}$ $\displaystyle 2 a b + 4 b^{2}$

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

$a^{2}$ $\displaystyle a^{2}$

Correct answer:

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 2 b\right)^{2} = \left( a + 2*b \right) \cdot \left( a + 2*b \right)$ $\displaystyle \left(a + 2 b\right)^{2} = \left( a + 2*b \right) \cdot \left( a + 2*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 2 b = 2 a b$ $\displaystyle a \cdot 2 b = 2 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 2 b = 2 a b$ $\displaystyle a \cdot 2 b = 2 a b$

Now we multiply the last terms of each expression together.

$2 b \cdot 2 b = 4 b^{2}$ $\displaystyle 2 b \cdot 2 b = 4 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

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Example Question #171 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

$\left(a + 17 b\right)^{2}$ $\displaystyle \left(a + 17 b\right)^{2}$

Possible Answers:

$a^{2} + 34 a b$ $\displaystyle a^{2} + 34 a b$

$a^{2}$ $\displaystyle a^{2}$

$a^{2} + 34 a b + 289 b^{2}$ $\displaystyle a^{2} + 34 a b + 289 b^{2}$

$17 a b + 289 b^{2}$ $\displaystyle 17 a b + 289 b^{2}$

$b^{2}$ $\displaystyle b^{2}$

Correct answer:

$a^{2} + 34 a b + 289 b^{2}$ $\displaystyle a^{2} + 34 a b + 289 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 17 b\right)^{2} = \left( a + 17*b \right) \cdot \left( a + 17*b \right)$ $\displaystyle \left(a + 17 b\right)^{2} = \left( a + 17*b \right) \cdot \left( a + 17*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 17 b = 17 a b$ $\displaystyle a \cdot 17 b = 17 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 17 b = 17 a b$ $\displaystyle a \cdot 17 b = 17 a b$

Now we multiply the last terms of each expression together.

$17 b \cdot 17 b = 289 b^{2}$ $\displaystyle 17 b \cdot 17 b = 289 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 34 a b + 289 b^{2}$ $\displaystyle a^{2} + 34 a b + 289 b^{2}$

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Example Question #4 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 3 b\right)^{2}$ $\displaystyle \left(a + 3 b\right)^{2}$

Possible Answers:

$a^{2} + 6 a b + 9 b^{2}$ $\displaystyle a^{2} + 6 a b + 9 b^{2}$

$a^{2}$ $\displaystyle a^{2}$

$a^{2} + 6 a b$ $\displaystyle a^{2} + 6 a b$

$3 a b + 9 b^{2}$ $\displaystyle 3 a b + 9 b^{2}$

$b^{2}$ $\displaystyle b^{2}$

Correct answer:

$a^{2} + 6 a b + 9 b^{2}$ $\displaystyle a^{2} + 6 a b + 9 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 3 b\right)^{2} = \left( a + 3*b \right) \cdot \left( a + 3*b \right)$ $\displaystyle \left(a + 3 b\right)^{2} = \left( a + 3*b \right) \cdot \left( a + 3*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 3 b = 3 a b$ $\displaystyle a \cdot 3 b = 3 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 3 b = 3 a b$ $\displaystyle a \cdot 3 b = 3 a b$

Now we multiply the last terms of each expression together.

$3 b \cdot 3 b = 9 b^{2}$ $\displaystyle 3 b \cdot 3 b = 9 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 6 a b + 9 b^{2}$ $\displaystyle a^{2} + 6 a b + 9 b^{2}$

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Example Question #171 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

$\left(a + 2 b\right)^{2}$ $\displaystyle \left(a + 2 b\right)^{2}$

Possible Answers:

$2 a b + 4 b^{2}$ $\displaystyle 2 a b + 4 b^{2}$

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

$a^{2}$ $\displaystyle a^{2}$

$a^{2} + 4 a b$ $\displaystyle a^{2} + 4 a b$

$b^{2}$ $\displaystyle b^{2}$

Correct answer:

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

Explanation:

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 2 b = 2 a b$ $\displaystyle a \cdot 2 b = 2 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 2 b = 2 a b$ $\displaystyle a \cdot 2 b = 2 a b$

Now we multiply the last terms of each expression together.

$2 b \cdot 2 b = 4 b^{2}$ $\displaystyle 2 b \cdot 2 b = 4 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 4 a b + 4 b^{2}$ $\displaystyle a^{2} + 4 a b + 4 b^{2}$

Report an Error

Example Question #5 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 15 b\right)^{2}$ $\displaystyle \left(a + 15 b\right)^{2}$

Possible Answers:

$15 a b + 225 b^{2}$ $\displaystyle 15 a b + 225 b^{2}$

$a^{2} + 30 a b + 225 b^{2}$ $\displaystyle a^{2} + 30 a b + 225 b^{2}$

$a^{2} + 30 a b$ $\displaystyle a^{2} + 30 a b$

$b^{2}$ $\displaystyle b^{2}$

$a^{2}$ $\displaystyle a^{2}$

Correct answer:

$a^{2} + 30 a b + 225 b^{2}$ $\displaystyle a^{2} + 30 a b + 225 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 15 b\right)^{2} = \left( a + 15*b \right) \cdot \left( a + 15*b \right)$ $\displaystyle \left(a + 15 b\right)^{2} = \left( a + 15*b \right) \cdot \left( a + 15*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 15 b = 15 a b$ $\displaystyle a \cdot 15 b = 15 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 15 b = 15 a b$ $\displaystyle a \cdot 15 b = 15 a b$

Now we multiply the last terms of each expression together.

$15 b \cdot 15 b = 225 b^{2}$ $\displaystyle 15 b \cdot 15 b = 225 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 30 a b + 225 b^{2}$ $\displaystyle a^{2} + 30 a b + 225 b^{2}$

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Example Question #243 : High School: Algebra

Use FOIL for the following expression.

$\left(a + 7 b\right)^{2}$ $\displaystyle \left(a + 7 b\right)^{2}$

Possible Answers:

$b^{2}$ $\displaystyle b^{2}$

$a^{2}$ $\displaystyle a^{2}$

$7 a b + 49 b^{2}$ $\displaystyle 7 a b + 49 b^{2}$

$a^{2} + 14 a b$ $\displaystyle a^{2} + 14 a b$

$a^{2} + 14 a b + 49 b^{2}$ $\displaystyle a^{2} + 14 a b + 49 b^{2}$

Correct answer:

$a^{2} + 14 a b + 49 b^{2}$ $\displaystyle a^{2} + 14 a b + 49 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 7 b\right)^{2} = \left( a + 7*b \right) \cdot \left( a + 7*b \right)$ $\displaystyle \left(a + 7 b\right)^{2} = \left( a + 7*b \right) \cdot \left( a + 7*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 7 b = 7 a b$ $\displaystyle a \cdot 7 b = 7 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 7 b = 7 a b$ $\displaystyle a \cdot 7 b = 7 a b$

Now we multiply the last terms of each expression together.

$7 b \cdot 7 b = 49 b^{2}$ $\displaystyle 7 b \cdot 7 b = 49 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 14 a b + 49 b^{2}$ $\displaystyle a^{2} + 14 a b + 49 b^{2}$

Report an Error

Example Question #6 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 6 b\right)^{2}$ $\displaystyle \left(a + 6 b\right)^{2}$

Possible Answers:

$6 a b + 36 b^{2}$ $\displaystyle 6 a b + 36 b^{2}$

$a^{2} + 12 a b$ $\displaystyle a^{2} + 12 a b$

$b^{2}$ $\displaystyle b^{2}$

$a^{2}$ $\displaystyle a^{2}$

$a^{2} + 12 a b + 36 b^{2}$ $\displaystyle a^{2} + 12 a b + 36 b^{2}$

Correct answer:

$a^{2} + 12 a b + 36 b^{2}$ $\displaystyle a^{2} + 12 a b + 36 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 6 b\right)^{2} = \left( a + 6*b \right) \cdot \left( a + 6*b \right)$ $\displaystyle \left(a + 6 b\right)^{2} = \left( a + 6*b \right) \cdot \left( a + 6*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 6 b = 6 a b$ $\displaystyle a \cdot 6 b = 6 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 6 b = 6 a b$ $\displaystyle a \cdot 6 b = 6 a b$

Now we multiply the last terms of each expression together.

$6 b \cdot 6 b = 36 b^{2}$ $\displaystyle 6 b \cdot 6 b = 36 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 12 a b + 36 b^{2}$ $\displaystyle a^{2} + 12 a b + 36 b^{2}$

Report an Error

Example Question #6 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 5 b\right)^{2}$ $\displaystyle \left(a + 5 b\right)^{2}$

Possible Answers:

$\ a^{2} + 10 a b$ $\displaystyle \ a^{2} + 10 a b$

$a^{2}$ $\displaystyle a^{2}$

$5 a b + 25 b^{2}$ $\displaystyle 5 a b + 25 b^{2}$

$b^{2}$ $\displaystyle b^{2}$

$a^{2} + 10 a b + 25 b^{2}$ $\displaystyle a^{2} + 10 a b + 25 b^{2}$

Correct answer:

$a^{2} + 10 a b + 25 b^{2}$ $\displaystyle a^{2} + 10 a b + 25 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 5 b\right)^{2} = \left( a + 5*b \right) \cdot \left( a + 5*b \right)$ $\displaystyle \left(a + 5 b\right)^{2} = \left( a + 5*b \right) \cdot \left( a + 5*b \right)$

Now we multiply the first parts of the first and second expression together.

$a \cdot a = a^{2}$ $\displaystyle a \cdot a = a^{2}$

Now we multiply the first term of the first expression with the second term of the second expression.

$a \cdot 5 b = 5 a b$ $\displaystyle a \cdot 5 b = 5 a b$

Now we multiply the second term of the first expression with the first term of the second expression.

$a \cdot 5 b = 5 a b$ $\displaystyle a \cdot 5 b = 5 a b$

Now we multiply the last terms of each expression together.

$5 b \cdot 5 b = 25 b^{2}$ $\displaystyle 5 b \cdot 5 b = 25 b^{2}$

Now we add all these results together, and we get.

$a^{2} + 10 a b + 25 b^{2}$ $\displaystyle a^{2} + 10 a b + 25 b^{2}$

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Polynomial Identities and Numerical Relationships: CCSS.Math.Content.HSA-APR.C.4

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #1 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #2 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #3 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #171 : Arithmetic With Polynomials & Rational Expressions

Example Question #4 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #171 : Arithmetic With Polynomials & Rational Expressions

Example Question #5 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #243 : High School: Algebra

Example Question #6 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #6 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

All Common Core: High School - Algebra Resources

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