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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

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Example Questions

Example Question #241 : High School: Algebra

What are the -intercept(s) of the function?

y=x^2+10x+25

Possible Answers:

$\text{x-intercepts: }x=-5 \text{ and } x=-1$

$\text{x-intercept: }x=-5$

$\text{x-intercepts: }x=-5 \text{ and } x=5$

$\text{x-intercepts: }x=-5 \text{ and } x=0$

$\text{x-intercept: }x=5$

Correct answer:

$\text{x-intercept: }x=-5$

Explanation:

To find the -intercept of a function, first recall that the -intercept represents the points where the graph of the function crosses the -axis. In other words where the function has a value equal to zero.

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+10x+25

the coefficients are,

$\\b=10 \\c=25$

therefore the factors of that have a sum of are,

y=(x+5)(x+5)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x+5=0\rightarrow x=-5$

To verify, graph the function.

Screen shot 2016 03 09 at 10.24.54 am

The graph crosses the -axis at -5, thus verifying the result found by factorization.

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

Use FOIL for the following expression.

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

Possible Answers:

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

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

$b^{2}$

$a^{2} + 20 a b$

$a^{2}$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 10 b = 10 a b$

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

$a \cdot 10 b = 10 a b$

Now we multiply the last terms of each expression together.

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

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

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

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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.

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

Possible Answers:

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

$a^{2} + 28 a b$

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

$a^{2}$

$b^{2}$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 14 b = 14 a b$

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

$a \cdot 14 b = 14 a b$

Now we multiply the last terms of each expression together.

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

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

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

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Example Question #242 : High School: Algebra

Use FOIL for the following expression.

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

Possible Answers:

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

$a^{2}$

$b^{2}$

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

$a^{2} + 4 a b$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 2 b = 2 a b$

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

$a \cdot 2 b = 2 a b$

Now we multiply the last terms of each expression together.

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

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

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

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

Use FOIL for the following expression.

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

Possible Answers:

$a^{2}$

$a^{2} + 34 a b$

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

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

$b^{2}$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 17 b = 17 a b$

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

$a \cdot 17 b = 17 a b$

Now we multiply the last terms of each expression together.

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

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

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

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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.

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

Possible Answers:

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

$b^{2}$

$a^{2} + 6 a b$

$a^{2}$

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

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 3 b = 3 a b$

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

$a \cdot 3 b = 3 a b$

Now we multiply the last terms of each expression together.

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

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

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

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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.

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

Possible Answers:

$a^{2} + 4 a b$

$b^{2}$

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

$a^{2}$

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

Correct answer:

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

Explanation:

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

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

$a \cdot 2 b = 2 a b$

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

$a \cdot 2 b = 2 a b$

Now we multiply the last terms of each expression together.

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

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

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

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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.

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

Possible Answers:

$a^{2}$

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

$a^{2} + 30 a b$

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

$b^{2}$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 15 b = 15 a b$

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

$a \cdot 15 b = 15 a b$

Now we multiply the last terms of each expression together.

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

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

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

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Example Question #242 : High School: Algebra

Use FOIL for the following expression.

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

Possible Answers:

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

$a^{2}$

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

$b^{2}$

$a^{2} + 14 a b$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 7 b = 7 a b$

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

$a \cdot 7 b = 7 a b$

Now we multiply the last terms of each expression together.

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

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

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

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

Use FOIL for the following expression.

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

Possible Answers:

$a^{2}$

$b^{2}$

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

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

$a^{2} + 12 a b$

Correct answer:

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

Explanation:

The first step is to rewrite the problem as follows.

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

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

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

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

$a \cdot 6 b = 6 a b$

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

$a \cdot 6 b = 6 a b$

Now we multiply the last terms of each expression together.

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

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

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

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All Common Core: High School - Algebra Resources

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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 #241 : High School: Algebra

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 #242 : High School: Algebra

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

Example Question #4 : 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 #242 : High School: Algebra

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

All Common Core: High School - Algebra Resources

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