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Algebra 1 : Variables

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Example Questions

Example Question #1 : How To Find The Value Of The Coefficient

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 2x+ 0.5\right )^{8}$ .

Possible Answers:

1,680

26,880

224

Correct answer:

224

Explanation:

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

A = 2, B =0.5, k=5, n = 8 :

$C(8,5) \cdot 2^{5} \cdot 0.5 ^{8-5}$

$=C(8,5) \cdot 2^{5} \cdot 0.5 ^{3}$

$= 56\cdot 32 \cdot 0.125$

=224

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Example Question #1 : How To Find The Value Of The Coefficient

Give the coefficient of $x^{4}$ in the binomial expansion of $\left (6x+ \frac{1}{6}\right )^{7}$ .

Possible Answers:

$\frac{35}{6}$

140

210

5,040

Correct answer:

210

Explanation:

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{4}$ coefficient can be determined by setting

$A = 6, B = \frac{1}{6}, n = 7, k= 4$ :

$C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{7-4}$

$=C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{3}$

$=35 \cdot 1,296 \cdot \frac{1}{216}$

= 210

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Example Question #3 : How To Find The Value Of The Coefficient

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 0.2x+ 5 \right )^{7}$ .

Possible Answers:

20.16

2,625

315,000

0.168

Correct answer:

0.168

Explanation:

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

A = 0.2, B =5, k=5, n = 7

$C(7,5) \cdot 0.2^{5} \cdot 5 ^{7 - 5}$

$= C(7,5) \cdot 0.2 ^{5} \cdot 5 ^{2}$

$= 21\cdot 0.00032 \cdot 25$

= 0.168

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Example Question #2 : How To Find The Value Of The Coefficient

Give the coefficient of $x^{2}$ in the product

$\left (3x- 7 \right ) \left ( 4x+3\right )\left ( 2x-7\right )$ .

Possible Answers:

158

Correct answer:

Explanation:

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left (\underline{3x}- 7 \right ) \left ( \underline{4x}+3\right )\left ( 2x\underline{-7}\right )$

$3x \cdot 4x \cdot (-7) = -84x^{2}$

$\left (\underline{3x}- 7 \right ) \left ( 4x\underline{+3}\right )\left ( \underline{2x}-7\right )$

$3x \cdot 3 \cdot 2x = 18x^{2}$

$\left (3x\underline{- 7} \right ) \left ( \underline{4x}+3\right )\left ( \underline{2x}-7\right )$

$-7 \cdot 4x \cdot 2x = -56x^{2}$

Add: $-84x^{2} + 18x^{2} - 56x^{2} = -122x^{2}$

The correct response is -122.

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Example Question #1 : How To Find The Value Of The Coefficient

What is the value of the coefficient of ? $\frac{3x^2-9x}{x}-4x$

Possible Answers:

Correct answer:

Explanation:

In order to determine the coefficient, we will need to fully simplify this expression.

The numerator of the first term shares an variable, which can be divided.

$\frac{3x^2-9x}{x} = \frac{x(3x-9)}{x} = 3x-9$

Subtract this expression with .

3x-9-4x = -x-9

The coefficient is the number in front of . The coefficient is .

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Example Question #1 : How To Find The Solution To A Binomial Problem

Solve for : (x+5)+(3x+15)=0

Possible Answers:

$\frac{20}{3}$

Correct answer:

Explanation:

First, rearrange the equation so that "like terms" are grouped together, like this: 1x+3x+5+15 .

Second, combine "like" terms with the appropriate mathematical function (i.e., addition, subtraction, etc.), so in this problem, you'll be left with 4x+20 .

Third, set the entire equation equal to and solve for :

4x+20=0

4x=-20

$\frac{4x}{4}=\frac{-20}{4}$

x=-5

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Example Question #2 : How To Find The Solution To A Binomial Problem

Solve for : (x-3)+(3x-2)=7 .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

First, rearrange the equation so that "like terms" are grouped together, like this: 1x+3x-3-2 .

Second, combine "like" terms with the appropriate mathematical function (i.e., addition, subtraction, etc.), so in this problem, you'll be left with

4x-5=7 .

Third, solve for :

4x-5=7

4x=12

$\frac{4x}{4}=\frac{12}{4}$

x=3

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Example Question #2 : Finding Roots

Solve for :

(x+3)^2=16

Possible Answers:

x=1

$x=4\ or\ x=-4$

x=-4

$x=1\ or\ x=-7$

Correct answer:

$x=1\ or\ x=-7$

Explanation:

(x+3)^2=16

$\sqrt{(x+3)^2}=\sqrt{16}\textup{}$

$x+3=\pm4$

$x=1\ or\ -7$

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Example Question #1 : How To Find The Solution To A Binomial Problem

4x-8y=32

Solve for in terms of .

Possible Answers:

x=8-2y

y=4-0.5x

x = 8 + 2y

y=0.5x-4

Correct answer:

x = 8 + 2y

Explanation:

First, add 8y to both sides of the equation, cancelling out -8y and isolating x to a value in terms of y:

4x - 8y = 32

+8y +8y

4x = 32 + 8y

Then divide both sides of the equation by 4, providing the x-value in terms of y:

x = 8 + 2y

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Example Question #3 : How To Find The Solution To A Binomial Problem

Solve in terms of :

$\frac{2}{3}x+5y=7$

Possible Answers:

$x=7+\frac{10}{3}y$

$x=7-\frac{10}{3}y$

$x=\frac{21}{2}-\frac{15}{2}y$

$x=\frac{21}{2}+\frac{15}{2}y$

Correct answer:

$x=\frac{21}{2}-\frac{15}{2}y$

Explanation:

First, isolate X by itself:

$\frac{2}{3}x+5y=7$

-5y -5y

$\frac{2}{3}x=7-5y$

Next, divide both sides by 2/3 to get X by itself:

$\frac{2}{3}\div\frac{2}{3}x=\frac{2}{3}\div\frac{7}{1}-\frac{2}{3}\div\frac{5}{1}y$

$\frac{3}{2}\cdot\frac{2}{3}x=\frac{3}{2}\cdot\frac{7}{1}-\frac{3}{2}\cdot\frac{5}{1}y$

Your answer should come out as:

$x=\frac{21}{2}-\frac{15}{2}y$

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Algebra 1 : Variables

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : How To Find The Value Of The Coefficient

Example Question #1 : How To Find The Value Of The Coefficient

Example Question #3 : How To Find The Value Of The Coefficient

Example Question #2 : How To Find The Value Of The Coefficient

Example Question #1 : How To Find The Value Of The Coefficient

Example Question #1 : How To Find The Solution To A Binomial Problem

Example Question #2 : How To Find The Solution To A Binomial Problem

Example Question #2 : Finding Roots

Example Question #1 : How To Find The Solution To A Binomial Problem

Example Question #3 : How To Find The Solution To A Binomial Problem

All Algebra 1 Resources

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