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

Use FOIL for the following expression.

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

Possible Answers:

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

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

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

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

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

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.

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

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Example Question #131 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

$\left(a + 18 b\right)^{2}$ $\displaystyle \left(a + 18 b\right)^{2}$

Which of the following expressions is equivalent to the expression above?

Possible Answers:

$18 a b + 324 b^{2}$ $\displaystyle 18 a b + 324 b^{2}$

$a^{2} + 36 a b + 324 b^{2}$ $\displaystyle a^{2} + 36 a b + 324 b^{2}$

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

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

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

Correct answer:

$a^{2} + 36 a b + 324 b^{2}$ $\displaystyle a^{2} + 36 a b + 324 b^{2}$

Explanation:

The first step is to rewrite the problem as follows.

$\left(a + 18 b\right)^{2} = \left( a + 18*b \right) \cdot \left( a + 18*b \right)$ $\displaystyle \left(a + 18 b\right)^{2} = \left( a + 18*b \right) \cdot \left( a + 18*b \right)$

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.

$a \cdot 18 b = 18 a b$ $\displaystyle a \cdot 18 b = 18 a b$

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

$a \cdot 18 b = 18 a b$ $\displaystyle a \cdot 18 b = 18 a b$

Now we multiply the last terms of each expression together.

$18 b \cdot 18 b = 324 b^{2}$ $\displaystyle 18 b \cdot 18 b = 324 b^{2}$

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

$a^{2} + 36 a b + 324 b^{2}$ $\displaystyle a^{2} + 36 a b + 324 b^{2}$

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

Example Question #131 : Polynomial Identities And Numerical Relationships: Ccss.Math.Content.Hsa Apr.C.4

All Common Core: High School - Algebra Resources

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