How to find the value of the coefficient - PSAT Math

Card 0 of 42

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

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

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

$= 21\cdot 0.00032 \cdot 25$

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

Tap to see back →

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

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

Therefore, the 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}$$

Give the coefficient of in the product

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

Tap to see back →

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: .

The correct response is .

Give the coefficient of in the product

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

Tap to see back →

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:

The correct response is -122.

Give the coefficient of in the product

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

Tap to see back →

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:

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

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

:

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

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

$= 56\cdot 32 \cdot 0.125$

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

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

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

$= 21\cdot 0.00032 \cdot 25$

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

Tap to see back →

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

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

Therefore, the 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}$$

Give the coefficient of in the product

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

Tap to see back →

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: .

The correct response is .

Give the coefficient of in the product

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

Tap to see back →

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:

The correct response is -122.

Give the coefficient of in the product

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

Tap to see back →

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:

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

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

:

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

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

$= 56\cdot 32 \cdot 0.125$

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

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

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

$= 21\cdot 0.00032 \cdot 25$

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

Tap to see back →

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

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

Therefore, the 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}$$

Give the coefficient of in the product

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

Tap to see back →

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: .

The correct response is .

Give the coefficient of in the product

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

Tap to see back →

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:

The correct response is -122.

Give the coefficient of in the product

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

Tap to see back →

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:

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

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

:

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

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

$= 56\cdot 32 \cdot 0.125$

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

Tap to see back →

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

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

Therefore, the coefficient can be determined by setting

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

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

$= 21\cdot 0.00032 \cdot 25$

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

Tap to see back →

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

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

Therefore, the 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}$$