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.

\(\displaystyle \frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}\)

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:

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

Cancel the \(\displaystyle x-2\) and cross multiply.

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

\(\displaystyle 4x+16=3x+21\)

\(\displaystyle x=5\)

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 \(\displaystyle -0.1\).

By multiplying with the foil method, we multiply our first values giving \(\displaystyle x^2\), our outside values giving \(\displaystyle 4x\). our inside values which gives \(\displaystyle -3x\), and out last values giving \(\displaystyle -12\).

We must begin by factoring out \(\displaystyle x^2\) from each term.

\(\displaystyle x^2(x^2-8x+12)\)

Next, we must find two numbers that sum to \(\displaystyle -8\) and multiply to \(\displaystyle 12\).

\(\displaystyle (-6)+(-2)=-8\)

\(\displaystyle -6*-2=12\)

Thus, our final answer is:

\(\displaystyle x^2(x-6)(x-2)\)