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PSAT Math : Binomials

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Example Questions

Example Question #1 : Binomials

Decrease 18d +30 by 40%. Which of the following will this be equal to?

Possible Answers:

18d -10

10.8d + 18

-22d + 40

10.8d + 30

18d +18

Correct answer:

10.8d + 18

Explanation:

A number decreased by 40% is equivalent to 100% of the number minus 40% of the number. This is taking 60% of the number, or, equivalently, multiplying it by 0.6.

Therefore, 18d +30 decreased by 40% is 0.6 times this, or

$0.6 \left (18d +30 \right ) = 0.6 \cdot 18d + 0.6 \cdot 30 = 10.8d + 18$

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Example Question #1 : Binomials

Find the product:

(7-2y)(5-6y)

Possible Answers:

12y^2-52y+35

12y^2-32y+35

52y+35

12y^2-35

12y^2-52y

Correct answer:

12y^2-52y+35

Explanation:

Find the product:

(7-2y)(5-6y)

Use the distributive property:

7(5-6y)-2y(5-6y)

35-42y-10y+12y^2

Write the resulting expression in standard form:

12y^2-52y+35

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Example Question #2 : Binomials

If 〖(x+y)〗²= 144 and 〖(x-y)〗²= 64, what is the value of xy?

Possible Answers:

Correct answer:

Explanation:

We first expand each binomial to get x² + 2xy + y² = 144 and x² - 2xy + y² = 64. We then subtract the second equation from the first to find 4xy = 80. Finally, we divide each side by 4 to find xy = 20.

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Example Question #1 : Binomials

Which of these expressions can be simplified further by collecting like terms?

Possible Answers:

7ab+13b

$7ab+12ab^{2}$

None of the expressions in the other choices can be simplified further

12a+12b

$7b+13b^{2}$

Correct answer:

None of the expressions in the other choices can be simplified further

Explanation:

A binomial can be simplified further if and only if the two terms have the same combination of variables and the same exponents for each like variable. This is not the case in any of the four binomials given, so none of the expressions can be simplified further.

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Example Question #3 : Binomials

Solve for .

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Possible Answers:

Correct answer:

Explanation:

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Factor the expression

numerator: find two numbers that add to 2 and multiply to -8 [use 4,-2]

denominator: find two numbers that add to 5 and multiply to -14 [use 7,-2]

new expression:

$\frac{(x+4)(x-2)}{(x+7)(x-2)}=\frac{3}{4}$

Cancel the x-2 and cross multiply.

4(x+4)=3(x+7)

4x+16=3x+21

x=5

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

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

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

Possible Answers:

$\frac{11}{12}$

$\frac{1}{4}$

$\frac{7}{6}$

$\frac{1}{2}$

$\frac{1}{6}$

Correct answer:

$\frac{1}{2}$

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:

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

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

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

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

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

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

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 #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:

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

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

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

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

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

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

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

The correct response is -122.

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PSAT Math : Binomials

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Binomials

Example Question #1 : Binomials

Example Question #2 : Binomials

Example Question #1 : Binomials

Example Question #3 : Binomials

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 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

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