How to find the value of the coefficient - Algebra

Card 0 of 63

Question

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

Answer

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

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

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

$= 21\cdot 0.00032 \cdot 25$

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Question

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

Answer

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

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

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

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

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

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Question

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

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

Answer

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 .

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Question

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

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

Answer

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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Question

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

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

Answer

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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Question

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

Answer

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

:

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

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

$= 56\cdot 32 \cdot 0.125$

Compare your answer with the correct one above

Question

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

Answer

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 .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$= 21\cdot 0.00032 \cdot 25$

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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 .

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

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

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

Answer

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 $\frac{1}{2}$ .

Compare your answer with the correct one above

Question

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

Answer

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

:

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

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

$= 56\cdot 32 \cdot 0.125$

Compare your answer with the correct one above

Question

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

Answer

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 .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

$= 21\cdot 0.00032 \cdot 25$

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

Compare your answer with the correct one above

Question

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

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

Answer

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 .

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

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

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

Answer

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 $\frac{1}{2}$ .

Compare your answer with the correct one above

Question

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

Answer

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

:

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

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

$= 56\cdot 32 \cdot 0.125$

Compare your answer with the correct one above

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