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

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 #71 : 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:

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

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

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

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

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

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.

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

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

$a\cdot5 b=5 a b$ $\displaystyle a\cdot5 b=5 a b$

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

$a\cdot5 b=5 a b$ $\displaystyle a\cdot5 b=5 a b$

Now we multiply the last terms of each expression together.

$5 b\cdot5 b=25 b^{2}$ $\displaystyle 5 b\cdot5 b=25 b^{2}$

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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Example Question #81 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

Report an Error

Example Question #251 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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

$a\cdot14 b=14 a b$ $\displaystyle a\cdot14 b=14 a b$

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

$a\cdot14 b=14 a b$ $\displaystyle a\cdot14 b=14 a b$

Now we multiply the last terms of each expression together.

$14 b\cdot14 b=196 b^{2}$ $\displaystyle 14 b\cdot14 b=196 b^{2}$

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

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

Report an Error

Example Question #251 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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

$a\cdot10 b=10 a b$ $\displaystyle a\cdot10 b=10 a b$

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

$a\cdot10 b=10 a b$ $\displaystyle a\cdot10 b=10 a b$

Now we multiply the last terms of each expression together.

$10 b\cdot10 b=100 b^{2}$ $\displaystyle 10 b\cdot10 b=100 b^{2}$

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

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

Report an Error

Example Question #252 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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

$a\cdot14 b=14 a b$ $\displaystyle a\cdot14 b=14 a b$

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

$a\cdot14 b=14 a b$ $\displaystyle a\cdot14 b=14 a b$

Now we multiply the last terms of each expression together.

$14 b\cdot14 b=196 b^{2}$ $\displaystyle 14 b\cdot14 b=196 b^{2}$

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

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

Report an Error

Example Question #81 : 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:

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

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

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

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

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

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.

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

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 #81 : 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:

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

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

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

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

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

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.

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

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

$a\cdot2 b=2 a b$ $\displaystyle a\cdot2 b=2 a b$

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

$a\cdot2 b=2 a b$ $\displaystyle a\cdot2 b=2 a b$

Now we multiply the last terms of each expression together.

$2 b\cdot2 b=4 b^{2}$ $\displaystyle 2 b\cdot2 b=4 b^{2}$

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 #254 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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 #255 : 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.

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

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

$a\cdot17 b=17 a b$ $\displaystyle a\cdot17 b=17 a b$

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

$a\cdot17 b=17 a b$ $\displaystyle a\cdot17 b=17 a b$

Now we multiply the last terms of each expression together.

$17 b\cdot17 b=289 b^{2}$ $\displaystyle 17 b\cdot17 b=289 b^{2}$

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

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

Report an Error

Example Question #256 : Arithmetic With Polynomials & Rational Expressions

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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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

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

Example Question #81 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #251 : Arithmetic With Polynomials & Rational Expressions

Example Question #251 : Arithmetic With Polynomials & Rational Expressions

Example Question #252 : Arithmetic With Polynomials & Rational Expressions

Example Question #81 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #81 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

Example Question #254 : Arithmetic With Polynomials & Rational Expressions

Example Question #255 : Arithmetic With Polynomials & Rational Expressions

Example Question #256 : Arithmetic With Polynomials & Rational Expressions

All Common Core: High School - Algebra Resources

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