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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

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #224 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

$\left(a + 8 b\right)^{2}$ $\displaystyle \left(a + 8 b\right)^{2}$

Possible Answers:

$a^{2} + 16 a b$ $\displaystyle a^{2} + 16 a b$

$8 a b + 64 b^{2}$ $\displaystyle 8 a b + 64 b^{2}$

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

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

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Correct answer:

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$ $\displaystyle \left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$

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

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

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

Now we multiply the last terms of each expression together.

$8 b\cdot8 b=64 b^{2}$ $\displaystyle 8 b\cdot8 b=64 b^{2}$

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

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

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Example Question #62 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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

$a\cdot7 b=7 a b$ $\displaystyle a\cdot7 b=7 a b$

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

$a\cdot7 b=7 a b$ $\displaystyle a\cdot7 b=7 a b$

Now we multiply the last terms of each expression together.

$7 b\cdot7 b=49 b^{2}$ $\displaystyle 7 b\cdot7 b=49 b^{2}$

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

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

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Example Question #231 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

$\left(a + 16 b\right)^{2}$ $\displaystyle \left(a + 16 b\right)^{2}$

Possible Answers:

$a^{2} + 32 a b$ $\displaystyle a^{2} + 32 a b$

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

$a^{2} + 32 a b + 256 b^{2}$ $\displaystyle a^{2} + 32 a b + 256 b^{2}$

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

$16 a b + 256 b^{2}$ $\displaystyle 16 a b + 256 b^{2}$

Correct answer:

$a^{2} + 32 a b + 256 b^{2}$ $\displaystyle a^{2} + 32 a b + 256 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 16 b\right)^{2}= \left(a + 16*b\right) \cdot \left(a + 16*b\right)$ $\displaystyle \left(a + 16 b\right)^{2}= \left(a + 16*b\right) \cdot \left(a + 16*b\right)$

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

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

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

$a\cdot16 b=16 a b$ $\displaystyle a\cdot16 b=16 a b$

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

$a\cdot16 b=16 a b$ $\displaystyle a\cdot16 b=16 a b$

Now we multiply the last terms of each expression together.

$16 b\cdot16 b=256 b^{2}$ $\displaystyle 16 b\cdot16 b=256 b^{2}$

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

$a^{2} + 32 a b + 256 b^{2}$ $\displaystyle a^{2} + 32 a b + 256 b^{2}$

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Example Question #232 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

$\left(a + 12 b\right)^{2}$ $\displaystyle \left(a + 12 b\right)^{2}$

Possible Answers:

$a^{2} + 24 a b$ $\displaystyle a^{2} + 24 a b$

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

$a^{2} + 24 a b + 144 b^{2}$ $\displaystyle a^{2} + 24 a b + 144 b^{2}$

$12 a b + 144 b^{2}$ $\displaystyle 12 a b + 144 b^{2}$

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

Correct answer:

$a^{2} + 24 a b + 144 b^{2}$ $\displaystyle a^{2} + 24 a b + 144 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 12 b\right)^{2}= \left(a + 12*b\right) \cdot \left(a + 12*b\right)$ $\displaystyle \left(a + 12 b\right)^{2}= \left(a + 12*b\right) \cdot \left(a + 12*b\right)$

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

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

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

$a\cdot12 b=12 a b$ $\displaystyle a\cdot12 b=12 a b$

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

$a\cdot12 b=12 a b$ $\displaystyle a\cdot12 b=12 a b$

Now we multiply the last terms of each expression together.

$12 b\cdot12 b=144 b^{2}$ $\displaystyle 12 b\cdot12 b=144 b^{2}$

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

$a^{2} + 24 a b + 144 b^{2}$ $\displaystyle a^{2} + 24 a b + 144 b^{2}$

Report an Error

Example Question #65 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 8 b\right)^{2}$ $\displaystyle \left(a + 8 b\right)^{2}$

Possible Answers:

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

$8 a b + 64 b^{2}$ $\displaystyle 8 a b + 64 b^{2}$

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

$a^{2} + 16 a b$ $\displaystyle a^{2} + 16 a b$

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

Correct answer:

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$ $\displaystyle \left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$

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

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

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

Now we multiply the last terms of each expression together.

$8 b\cdot8 b=64 b^{2}$ $\displaystyle 8 b\cdot8 b=64 b^{2}$

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

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Report an Error

Example Question #61 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a\cdotb=a b$ $\displaystyle a\cdotb=a b$

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

$a\cdotb=a b$ $\displaystyle a\cdotb=a b$

Now we multiply the last terms of each expression together.

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

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

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

Report an Error

Example Question #67 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 8 b\right)^{2}$ $\displaystyle \left(a + 8 b\right)^{2}$

Possible Answers:

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

$8 a b + 64 b^{2}$ $\displaystyle 8 a b + 64 b^{2}$

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

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

$a^{2} + 16 a b$ $\displaystyle a^{2} + 16 a b$

Correct answer:

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$ $\displaystyle \left(a + 8 b\right)^{2}= \left(a + 8*b\right) \cdot \left(a + 8*b\right)$

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

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

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

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

$a\cdot8 b=8 a b$ $\displaystyle a\cdot8 b=8 a b$

Now we multiply the last terms of each expression together.

$8 b\cdot8 b=64 b^{2}$ $\displaystyle 8 b\cdot8 b=64 b^{2}$

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

$a^{2} + 16 a b + 64 b^{2}$ $\displaystyle a^{2} + 16 a b + 64 b^{2}$

Report an Error

Example Question #68 : 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:

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

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

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

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

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

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.

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

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

$a\cdot15 b=15 a b$ $\displaystyle a\cdot15 b=15 a b$

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

$a\cdot15 b=15 a b$ $\displaystyle a\cdot15 b=15 a b$

Now we multiply the last terms of each expression together.

$15 b\cdot15 b=225 b^{2}$ $\displaystyle 15 b\cdot15 b=225 b^{2}$

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

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

Report an Error

Example Question #69 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 19 b\right)^{2}$ $\displaystyle \left(a + 19 b\right)^{2}$

Possible Answers:

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

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

$19 a b + 361 b^{2}$ $\displaystyle 19 a b + 361 b^{2}$

$a^{2} + 38 a b + 361 b^{2}$ $\displaystyle a^{2} + 38 a b + 361 b^{2}$

$a^{2} + 38 a b$ $\displaystyle a^{2} + 38 a b$

Correct answer:

$a^{2} + 38 a b + 361 b^{2}$ $\displaystyle a^{2} + 38 a b + 361 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 19 b\right)^{2}= \left(a + 19*b\right) \cdot \left(a + 19*b\right)$ $\displaystyle \left(a + 19 b\right)^{2}= \left(a + 19*b\right) \cdot \left(a + 19*b\right)$

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

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

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

$a\cdot19 b=19 a b$ $\displaystyle a\cdot19 b=19 a b$

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

$a\cdot19 b=19 a b$ $\displaystyle a\cdot19 b=19 a b$

Now we multiply the last terms of each expression together.

$19 b\cdot19 b=361 b^{2}$ $\displaystyle 19 b\cdot19 b=361 b^{2}$

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

$a^{2} + 38 a b + 361 b^{2}$ $\displaystyle a^{2} + 38 a b + 361 b^{2}$

Report an Error

Example Question #70 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

$\left(a + 9 b\right)^{2}$ $\displaystyle \left(a + 9 b\right)^{2}$

Possible Answers:

$a^{2} + 18 a b$ $\displaystyle a^{2} + 18 a b$

$9 a b + 81 b^{2}$ $\displaystyle 9 a b + 81 b^{2}$

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

$a^{2} + 18 a b + 81 b^{2}$ $\displaystyle a^{2} + 18 a b + 81 b^{2}$

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

Correct answer:

$a^{2} + 18 a b + 81 b^{2}$ $\displaystyle a^{2} + 18 a b + 81 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 9 b\right)^{2}= \left(a + 9*b\right) \cdot \left(a + 9*b\right)$ $\displaystyle \left(a + 9 b\right)^{2}= \left(a + 9*b\right) \cdot \left(a + 9*b\right)$

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

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

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

$a\cdot9 b=9 a b$ $\displaystyle a\cdot9 b=9 a b$

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

$a\cdot9 b=9 a b$ $\displaystyle a\cdot9 b=9 a b$

Now we multiply the last terms of each expression together.

$9 b\cdot9 b=81 b^{2}$ $\displaystyle 9 b\cdot9 b=81 b^{2}$

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

$a^{2} + 18 a b + 81 b^{2}$ $\displaystyle a^{2} + 18 a b + 81 b^{2}$

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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 #224 : Arithmetic With Polynomials & Rational Expressions

Example Question #62 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #231 : Arithmetic With Polynomials & Rational Expressions

Example Question #232 : Arithmetic With Polynomials & Rational Expressions

Example Question #65 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #61 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #67 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #68 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #69 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #70 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

All Common Core: High School - Algebra Resources

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