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ACT Math : Coefficients

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Example Questions

Example Question #1 : Coefficients

What is the value of the coefficient in front of the term that includes $x^{2}y^{7}$ in the expansion of $\left ( 2x-y \right )^{9}$ ?

Possible Answers:

144

Correct answer:

Explanation:

Using the binomial theorem, the term containing the x² y⁷will be equal to

(2x)²(–y)⁷

=36(–4x² y⁷)= -144x²y⁷

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Example Question #2 : Coefficients

A function of the form f(x)=ax^2+b passes through the points (0,7) and (-2,19) . What is the value of ?

Possible Answers:

Correct answer:

Explanation:

The easisest way to solve for is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in for and for f(x) . This gives us the equation

7=a(0)^2+b .

Squaring 0 gives us 0, and multiplying this by still gives 0, leaving only on the right side, such that

7 = b .

We now know the value of , and we can use this to help us find . Substituting our second set of coordinates into the function, we get

19 = a(-2)^2+b

which simplifies to

19=4a+b .

However, since we know b=7 , we can substitute to get

19=4a+7

subtracting 7 from both sides gives

12 = 4a

and dividing by 4 gives our answer

3 = a .

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

$2x^{2} \cdot x^{3}y^{2} \cdot 3y$ is equivalent to which of the following?

Possible Answers:

$5x^{6}y^{2}$

$6x^{5}y^{3}$

$5x^{5}y^{3}$

6xy

$6x^{6}y^{2}$

Correct answer:

$6x^{5}y^{3}$

Explanation:

To answer this problem, we need to multiply the expressions together, being mindful of how to correctly multiply like variables with exponents. To do this, we add the exponents together if the the like variables are being multiplied and subtract the exponents if the variables are being divided. So, for the presented data:

$2x^{2} \cdot x^{3}y^{2} \cdot 3y = 2x^{5} \cdot 3y^{3}$

We then multiply the remaining expressions together. When we do this, we will multiply the coefficients together and combine the different variables into the final expression. Therefore:

$2x^{5} \cdot 3y^{3} = 6x^{5}y^{3}$

This means our answer is $6x^{5}y^{3}$ .

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Example Question #4521 : Algebra 1

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

$\left ( x+ 0.4\right ) (x - 0.2) (3x-0.7)$ .

Possible Answers:

1.3

2.5

0.5

0.7

-0.1

Correct answer:

-0.1

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:

$\left (\underline{ x}+ 0.4\right ) (\underline{x} - 0.2) (3x\underline{-0.7})$

$x \cdot x \cdot (-0.7) = -0.7x^{2}$

$\left (\underline{ x}+ 0.4\right ) (x \underline{-0.2})(\underline{3x}-0.7)$

$x \cdot (-0.2) \cdot 3x= -0.6x^{2}$

$\left (x+ \underline{0.4}\right ) (\underline{x} - 0.2)(\underline{3x}-0.7)$

$0.4 \cdot x \cdot 3x = 1.2 x^{2}$

Add: $-0.7x^{2}+ (-0.6x^{2})+ 1.2x^{2} = -0.1x^{2}$ .

The correct response is -0.1 .

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Example Question #4522 : Algebra 1

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

$\left ( 2x + \frac{1}{3}\right ) \left ( x- \frac{1}{6}\right ) \left ( x+ \frac{1}{4}\right )$

Possible Answers:

$\frac{1}{6}$

$\frac{11}{12}$

$\frac{7}{6}$

$\frac{1}{4}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( x\underline{+ \frac{1}{4}}\right )$

$2x \cdot x \cdot \frac{1}{4} = \frac{1}{2}x^{2}$

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( x \underline{- \frac{1}{6}}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

$2x \cdot \left (- \frac{1}{6} \right ) \cdot x = - \frac{1}{3}x^{2}$

$\left ( 2x\underline{ + \frac{1}{3}}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

$\frac{1}{3} \cdot x \cdot x= \frac{1}{3}x^{2}$

Add: $\frac{1}{2}x^{2} +\left ( - \frac{1}{3}x^{2} \right ) + \frac{1}{3}x^{2} = \frac{1}{2}x^{2}$

The correct response is $\frac{1}{2}$ .

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Example Question #2 : 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:

26,880

224

1,680

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:

5,040

140

210

$\frac{35}{6}$

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:

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

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

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

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

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

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

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

The correct response is -122.

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ACT Math : Coefficients

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Coefficients

Example Question #2 : Coefficients

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

Example Question #4521 : Algebra 1

Example Question #4522 : Algebra 1

Example Question #2 : 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

All ACT Math Resources

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