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 $\displaystyle x$ terms and one constant are multiplied; find the three products and add them, as follows:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle A = 2, B =0.5, k=5, n = 8$:

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

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

$\displaystyle = 56\cdot 32 \cdot 0.125$

$\displaystyle =224$

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 210$

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

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

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

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

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

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

$\displaystyle A = 0.2, B =5, k=5, n = 7$

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

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

$\displaystyle = 21\cdot 0.00032 \cdot 25$

$\displaystyle = 0.168$

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 $\displaystyle x$ terms and one constant are multiplied; find the three products and add them, as follows:

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

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

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

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

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

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

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

The correct response is -122.

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 $\displaystyle x$ terms and one constant are multiplied; find the three products and add them, as follows:

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

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

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

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

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

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

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

The correct response is $\displaystyle -0.1$.