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Common Core: High School - Algebra : Using the Binomial Theorem for Expansion: CCSS.Math.Content.HSA-APR.C.5

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Example Questions

Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand,

$\left(x + y\right)^{12}$

Possible Answers:

$x^{12} + 12 x^{11} y + 66 x^{10} y^{2}$

$\\ x^{12} + 12 x^{11} y + 66 x^{10} y^{2} + 220 x^{9} y^{3} + 495 x^{8} y^{4} + 792 x^{7} y^{5} + 924 x^{6} y^{6} + 792 x^{5} y^{7} + 495 x^{4} y^{8} + 220 x^{3} y^{9} + 66 x^{2} y^{10} + 12 x y^{11} + y^{12}$

$x^{12} + y^{12}$

$x^{12} + 12 x y^{11} + y^{11}$

Not Possible

Correct answer:

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 12 we can replace $\uptext{n}$ , with 12 .

Now our equation looks like

$\sum_{i=0}^{12} x^{- i + 12} y^{i} {\binom{12}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 12 }{ 0 } x ^{\left( 12 - 0 \right)} y ^{ 0 }= x^{12}$

Term 2 : $\binom{ 12 }{ 1 } x ^{\left( 12 - 1 \right)} y ^{ 1 }= 12 x^{11} y$

Term 3 : $\binom{ 12 }{ 2 } x ^{\left( 12 - 2 \right)} y ^{ 2 }= 66 x^{10} y^{2}$

Term 4 : $\binom{ 12 }{ 3 } x ^{\left( 12 - 3 \right)} y ^{ 3 }= 220 x^{9} y^{3}$

Term 5 : $\binom{ 12 }{ 4 } x ^{\left( 12 - 4 \right)} y ^{ 4 }= 495 x^{8} y^{4}$

Term 6 : $\binom{ 12 }{ 5 } x ^{\left( 12 - 5 \right)} y ^{ 5 }= 792 x^{7} y^{5}$

Term 7 : $\binom{ 12 }{ 6 } x ^{\left( 12 - 6 \right)} y ^{ 6 }= 924 x^{6} y^{6}$

Term 8 : $\binom{ 12 }{ 7 } x ^{\left( 12 - 7 \right)} y ^{ 7 }= 792 x^{5} y^{7}$

Term 9 : $\binom{ 12 }{ 8 } x ^{\left( 12 - 8 \right)} y ^{ 8 }= 495 x^{4} y^{8}$

Term 10 : $\binom{ 12 }{ 9 } x ^{\left( 12 - 9 \right)} y ^{ 9 }= 220 x^{3} y^{9}$

Term 11 : $\binom{ 12 }{ 10 } x ^{\left( 12 - 10 \right)} y ^{ 10 }= 66 x^{2} y^{10}$

Term 12 : $\binom{ 12 }{ 11 } x ^{\left( 12 - 11 \right)} y ^{ 11 }= 12 x y^{11}$

Term 13 : $\binom{ 12 }{ 12 } x ^{\left( 12 - 12 \right)} y ^{ 12 }= y^{12}$

Now we combine the expressions and we get

$\\x^{12} + 12 x^{11} y + 66 x^{10} y^{2} + 220 x^{9} y^{3} + 495 x^{8} y^{4} + 792 x^{7} y^{5} + 924 x^{6} y^{6} + 792 x^{5} y^{7} + 495 x^{4} y^{8} + 220 x^{3} y^{9} + 66 x^{2} y^{10} + 12 x y^{11} + y^{12}$

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Example Question #2 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{8} y^{9}$ in the expansion of $\left(x + y\right)^{17}$ ?

Possible Answers:

There is no coefficient

24310

Correct answer:

24310

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 17 and i = 9

Now we compute the following

${\binom{n}{i}} = \binom{ 17 }{ 9 } = 24310$

Report an Error

Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{18}$

Possible Answers:

Not Possible

$x^{18} + y^{18}$

$x^{18} + 18 x y^{17} + y^{17}$

$x^{18} + 18 x^{17} y + 153 x^{16} y^{2}$

$\\x^{18} + 18 x^{17} y + 153 x^{16} y^{2} + 816 x^{15} y^{3} + 3060 x^{14} y^{4} + 8568 x^{13} y^{5} + 18564 x^{12} y^{6} + 31824 x^{11} y^{7} + 43758 x^{10} y^{8} + 48620 x^{9} y^{9} + 43758 x^{8} y^{10} + 31824 x^{7} y^{11} + 18564 x^{6} y^{12} + 8568 x^{5} y^{13} + 3060 x^{4} y^{14} + 816 x^{3} y^{15} + 153 x^{2} y^{16} + 18 x y^{17} + y^{18}$

Correct answer:

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 18 we can replace $\uptext{n}$ , with 18 .

Now our equation looks like

$\sum_{i=0}^{18} x^{- i + 18} y^{i} {\binom{18}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 18 }{ 0 } x ^{\left( 18 - 0 \right)} y ^{ 0 }= x^{18}$

Term 2 : $\binom{ 18 }{ 1 } x ^{\left( 18 - 1 \right)} y ^{ 1 }= 18 x^{17} y$

Term 3 : $\binom{ 18 }{ 2 } x ^{\left( 18 - 2 \right)} y ^{ 2 }= 153 x^{16} y^{2}$

Term 4 : $\binom{ 18 }{ 3 } x ^{\left( 18 - 3 \right)} y ^{ 3 }= 816 x^{15} y^{3}$

Term 5 : $\binom{ 18 }{ 4 } x ^{\left( 18 - 4 \right)} y ^{ 4 }= 3060 x^{14} y^{4}$

Term 6 : $\binom{ 18 }{ 5 } x ^{\left( 18 - 5 \right)} y ^{ 5 }= 8568 x^{13} y^{5}$

Term 7 : $\binom{ 18 }{ 6 } x ^{\left( 18 - 6 \right)} y ^{ 6 }= 18564 x^{12} y^{6}$

Term 8 : $\binom{ 18 }{ 7 } x ^{\left( 18 - 7 \right)} y ^{ 7 }= 31824 x^{11} y^{7}$

Term 9 : $\binom{ 18 }{ 8 } x ^{\left( 18 - 8 \right)} y ^{ 8 }= 43758 x^{10} y^{8}$

Term 10 : $\binom{ 18 }{ 9 } x ^{\left( 18 - 9 \right)} y ^{ 9 }= 48620 x^{9} y^{9}$

Term 11 : $\binom{ 18 }{ 10 } x ^{\left( 18 - 10 \right)} y ^{ 10 }= 43758 x^{8} y^{10}$

Term 12 : $\binom{ 18 }{ 11 } x ^{\left( 18 - 11 \right)} y ^{ 11 }= 31824 x^{7} y^{11}$

Term 13 : $\binom{ 18 }{ 12 } x ^{\left( 18 - 12 \right)} y ^{ 12 }= 18564 x^{6} y^{12}$

Term 14 : $\binom{ 18 }{ 13 } x ^{\left( 18 - 13 \right)} y ^{ 13 }= 8568 x^{5} y^{13}$

Term 15 : $\binom{ 18 }{ 14 } x ^{\left( 18 - 14 \right)} y ^{ 14 }= 3060 x^{4} y^{14}$

Term 16 : $\binom{ 18 }{ 15 } x ^{\left( 18 - 15 \right)} y ^{ 15 }= 816 x^{3} y^{15}$

Term 17 : $\binom{ 18 }{ 16 } x ^{\left( 18 - 16 \right)} y ^{ 16 }= 153 x^{2} y^{16}$

Term 18 : $\binom{ 18 }{ 17 } x ^{\left( 18 - 17 \right)} y ^{ 17 }= 18 x y^{17}$

Term 19 : $\binom{ 18 }{ 18 } x ^{\left( 18 - 18 \right)} y ^{ 18 }= y^{18}$

Now we combine the expressions and we get

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Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{6} y^{7}$ in the expansion of $\left(x + y\right)^{13}$ ?

Possible Answers:

There is no coefficient

1716

Correct answer:

1716

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of {y} , and the exponent of the original equation.

In this case n = 13 and i = 7

Now we compute the following

${\binom{n}{i}} = \binom{ 13 }{ 7 } = 1716$

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Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{8}$

Possible Answers:

$x^{8} + 8 x y^{7} + y^{7}$

Not Possible

$x^{8} + 8 x^{7} y + 28 x^{6} y^{2} + 56 x^{5} y^{3} + 70 x^{4} y^{4} + 56 x^{3} y^{5} + 28 x^{2} y^{6} + 8 x y^{7} + y^{8}$

$x^{8} + 8 x^{7} y + 28 x^{6} y^{2}$

$x^{8} + y^{8}$

Correct answer:

$x^{8} + 8 x^{7} y + 28 x^{6} y^{2} + 56 x^{5} y^{3} + 70 x^{4} y^{4} + 56 x^{3} y^{5} + 28 x^{2} y^{6} + 8 x y^{7} + y^{8}$

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 8 we can replace $\uptext{n}$ , with 8 .

Now our equation looks like

$\sum_{i=0}^{8} x^{- i + 8} y^{i} {\binom{8}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 8 }{ 0 } x ^{\left( 8 - 0 \right)} y ^{ 0 }= x^{8}$

Term 2 : $\binom{ 8 }{ 1 } x ^{\left( 8 - 1 \right)} y ^{ 1 }= 8 x^{7} y$

Term 3 : $\binom{ 8 }{ 2 } x ^{\left( 8 - 2 \right)} y ^{ 2 }= 28 x^{6} y^{2}$

Term 4 : $\binom{ 8 }{ 3 } x ^{\left( 8 - 3 \right)} y ^{ 3 }= 56 x^{5} y^{3}$

Term 5 : $\binom{ 8 }{ 4 } x ^{\left( 8 - 4 \right)} y ^{ 4 }= 70 x^{4} y^{4}$

Term 6 : $\binom{ 8 }{ 5 } x ^{\left( 8 - 5 \right)} y ^{ 5 }= 56 x^{3} y^{5}$

Term 7 : $\binom{ 8 }{ 6 } x ^{\left( 8 - 6 \right)} y ^{ 6 }= 28 x^{2} y^{6}$

Term 8 : $\binom{ 8 }{ 7 } x ^{\left( 8 - 7 \right)} y ^{ 7 }= 8 x y^{7}$

Term 9 : $\binom{ 8 }{ 8 } x ^{\left( 8 - 8 \right)} y ^{ 8 }= y^{8}$

Now we combine the expressions and we get

$x^{8} + 8 x^{7} y + 28 x^{6} y^{2} + 56 x^{5} y^{3} + 70 x^{4} y^{4} + 56 x^{3} y^{5} + 28 x^{2} y^{6} + 8 x y^{7} + y^{8}$

Report an Error

Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{9} y^{6}$ in the expansion of $\left(x + y\right)^{15}$ ?

Possible Answers:

5005

There is no coefficient

Correct answer:

5005

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 15 and i = 6

Now we compute the following

${\binom{n}{i}} = \binom{ 15 }{ 6 } = 5005$

Report an Error

Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{7}$

Possible Answers:

Not Possible

$x^{7} + y^{7}$

$x^{7} + 7 x^{6} y + 21 x^{5} y^{2}$

$x^{7} + 7 x^{6} y + 21 x^{5} y^{2} + 35 x^{4} y^{3} + 35 x^{3} y^{4} + 21 x^{2} y^{5} + 7 x y^{6} + y^{7}$

$x^{7} + 7 x y^{6} + y^{6}$

Correct answer:

$x^{7} + 7 x^{6} y + 21 x^{5} y^{2} + 35 x^{4} y^{3} + 35 x^{3} y^{4} + 21 x^{2} y^{5} + 7 x y^{6} + y^{7}$

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 7 we can replace $\uptext{n}$ , with 7 .

Now our equation looks like

$\sum_{i=0}^{7} x^{- i + 7} y^{i} {\binom{7}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 7 }{ 0 } x ^{\left( 7 - 0 \right)} y ^{ 0 }= x^{7}$

Term 2 : $\binom{ 7 }{ 1 } x ^{\left( 7 - 1 \right)} y ^{ 1 }= 7 x^{6} y$

Term 3 : $\binom{ 7 }{ 2 } x ^{\left( 7 - 2 \right)} y ^{ 2 }= 21 x^{5} y^{2}$

Term 4 : $\binom{ 7 }{ 3 } x ^{\left( 7 - 3 \right)} y ^{ 3 }= 35 x^{4} y^{3}$

Term 5 : $\binom{ 7 }{ 4 } x ^{\left( 7 - 4 \right)} y ^{ 4 }= 35 x^{3} y^{4}$

Term 6 : $\binom{ 7 }{ 5 } x ^{\left( 7 - 5 \right)} y ^{ 5 }= 21 x^{2} y^{5}$

Term 7 : $\binom{ 7 }{ 6 } x ^{\left( 7 - 6 \right)} y ^{ 6 }= 7 x y^{6}$

Term 8 : $\binom{ 7 }{ 7 } x ^{\left( 7 - 7 \right)} y ^{ 7 }= y^{7}$

Now we combine the expressions and we get

$x^{7} + 7 x^{6} y + 21 x^{5} y^{2} + 35 x^{4} y^{3} + 35 x^{3} y^{4} + 21 x^{2} y^{5} + 7 x y^{6} + y^{7}$

Report an Error

Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{11} y^{8}$ in the expansion of $\left(x + y\right)^{19}$ ?

Possible Answers:

75582

There is no coefficient

Correct answer:

75582

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 19 and i = 8

Now we compute the following

${\binom{n}{i}} = \binom{ 19 }{ 8 } = 75582$

Report an Error

Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Use Pascal's Triangle to Expand

$\left(x + y\right)^{14}$

Possible Answers:

$\\x^{14} + 14 x^{13} y + 91 x^{12} y^{2} + 364 x^{11} y^{3} + 1001 x^{10} y^{4} + 2002 x^{9} y^{5} + 3003 x^{8} y^{6} + 3432 x^{7} y^{7} + 3003 x^{6} y^{8} + 2002 x^{5} y^{9} + 1001 x^{4} y^{10} + 364 x^{3} y^{11} + 91 x^{2} y^{12} + 14 x y^{13} + y^{14}$

Not Possible

$x^{14} + 14 x y^{13} + y^{13}$

$x^{14} + 14 x^{13} y + 91 x^{12} y^{2}$

$x^{14} + y^{14}$

Correct answer:

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

Since the exponent in the question is 14 we can replace $\uptext{n}$ , with 14 .

Now our equation looks like

$\sum_{i=0}^{14} x^{- i + 14} y^{i} {\binom{14}{i}}$

Now we compute the sum, term by term.

Term 1 : $\binom{ 14 }{ 0 } x ^{\left( 14 - 0 \right)} y ^{ 0 }= x^{14}$

Term 2 : $\binom{ 14 }{ 1 } x ^{\left( 14 - 1 \right)} y ^{ 1 }= 14 x^{13} y$

Term 3 : $\binom{ 14 }{ 2 } x ^{\left( 14 - 2 \right)} y ^{ 2 }= 91 x^{12} y^{2}$

Term 4 : $\binom{ 14 }{ 3 } x ^{\left( 14 - 3 \right)} y ^{ 3 }= 364 x^{11} y^{3}$

Term 5 : $\binom{ 14 }{ 4 } x ^{\left( 14 - 4 \right)} y ^{ 4 }= 1001 x^{10} y^{4}$

Term 6 : $\binom{ 14 }{ 5 } x ^{\left( 14 - 5 \right)} y ^{ 5 }= 2002 x^{9} y^{5}$

Term 7 : $\binom{ 14 }{ 6 } x ^{\left( 14 - 6 \right)} y ^{ 6 }= 3003 x^{8} y^{6}$

Term 8 : $\binom{ 14 }{ 7 } x ^{\left( 14 - 7 \right)} y ^{ 7 }= 3432 x^{7} y^{7}$

Term 9 : $\binom{ 14 }{ 8 } x ^{\left( 14 - 8 \right)} y ^{ 8 }= 3003 x^{6} y^{8}$

Term 10 : $\binom{ 14 }{ 9 } x ^{\left( 14 - 9 \right)} y ^{ 9 }= 2002 x^{5} y^{9}$

Term 11 : $\binom{ 14 }{ 10 } x ^{\left( 14 - 10 \right)} y ^{ 10 }= 1001 x^{4} y^{10}$

Term 12 : $\binom{ 14 }{ 11 } x ^{\left( 14 - 11 \right)} y ^{ 11 }= 364 x^{3} y^{11}$

Term 13 : $\binom{ 14 }{ 12 } x ^{\left( 14 - 12 \right)} y ^{ 12 }= 91 x^{2} y^{12}$

Term 14 : $\binom{ 14 }{ 13 } x ^{\left( 14 - 13 \right)} y ^{ 13 }= 14 x y^{13}$

Term 15 : $\binom{ 14 }{ 14 } x ^{\left( 14 - 14 \right)} y ^{ 14 }= y^{14}$

Now we combine the expressions and we get

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Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

What is the coefficient of $x^{10} y^{4}$ in the expansion of $\left(x + y\right)^{14}$ ?

Possible Answers:

1001

There is no coefficient

Correct answer:

1001

Explanation:

In order to do this, we need to recall the formula for Pascal's Triangle.

$\sum_{i=0}^{n} x^{- i + n} y^{i} {\binom{n}{i}}$

The part in the expression that we care about is the combination.

We simply do this by looking at the exponent of $\uptext{y}$ , and the exponent of the original equation.

In this case n = 14 and i = 4

Now we compute the following

${\binom{n}{i}} = \binom{ 14 }{ 4 } = 1001$

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Common Core: High School - Algebra : Using the Binomial Theorem for Expansion: CCSS.Math.Content.HSA-APR.C.5

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Example Question #2 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

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Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

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Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5

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All Common Core: High School - Algebra Resources

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