First, FOIL:

\(\displaystyle \frac{1}{3}(8x^3-\frac{4}{3}x^2-24x+\frac{12}{3})\)

Simplify:

\(\displaystyle \frac{1}{3}(8x^3-\frac{4}{3}x^2-24x+4)\)

Distribute the \(\displaystyle \frac{1}{3}\) through the parentheses:

\(\displaystyle \frac{8}{3}x^3-\frac{4}{9}x^2-\frac{24}{3}x+\frac{4}{3}\)

Rewrite to make the expression look like one of the answer choices:

\(\displaystyle 2\frac{2}{3}x^3-\frac{4}{9}x^2-8x+1\frac{1}{3}\)

When simplifying polynomials, only combine the variables with like terms.

\(\displaystyle 15x^{3}y^{2}\) can be added to \(\displaystyle 3x^{3}y^{2}\), giving \(\displaystyle 18x^{3}y^{2}\). 

\(\displaystyle 4x^{2}\) can be subtracted from \(\displaystyle 8x^{2}\) to give \(\displaystyle 4x^{2}\).

Combine both of the terms into one expression to find the answer: \(\displaystyle 18x^{3}y^{2}+4x^{2}\)

\(\displaystyle (5c^{2}+5)+(-5c-5)\)

This is not a FOIL problem, as we are adding rather than multiplying the terms in parentheses.

Add like terms to solve.

\(\displaystyle 5c^{2}\) and \(\displaystyle -5c\) have no like terms and cannot be combined with anything.

5 and -5 can be combined however:

\(\displaystyle 5-5=0\)

This leaves us with \(\displaystyle 5c^{2}-5c\).

\(\displaystyle 4x^3-2x(x-2)(x+3)\)

First, FOIL the two binomials:

\(\displaystyle 4x^3-2x(x^2+x-6)\)

Then distribute the \(\displaystyle 2x\) through the terms in parentheses:

\(\displaystyle 4x^3-2x^3-2x^2+12x\)

Combine like terms:

\(\displaystyle 2x^3-2x^2+12x\)

This is not a FOIL problem, as we are adding rather than multiplying the terms in parentheses.

Add like terms together:

\(\displaystyle 2q^{2}-3q^{2}=-1q^{2}=-q^2\)

\(\displaystyle 4q\) has no like terms.

\(\displaystyle 7-(-5)=12\)

Combine these terms into one expression to find the answer:

\(\displaystyle -q^{2}+4q+12\)

When multiplying exponential components, you must add the powers of each term together.

\(\displaystyle 5+16=21\)

\(\displaystyle x^{5}*x^{16}=x^{21}\)

When multiplying polynomials, add the powers of each like-termed variable together.

For x: 5 + 2 = 7

For y: 17 + 2 = 19

Therefore the answer is \(\displaystyle x^{7}y^{19}\) .

\(\displaystyle \frac{x^2-10x+25}{x^2-25}\)

First we will factor the numerator:

\(\displaystyle x^2-10x+25=(x-5)(x-5)\)

Then factor the denominator:

\(\displaystyle x^2-25=(x+5)(x-5)\)

We can re-write the original fraction with these factors and then cancel an (x-5) term from both parts:

\(\displaystyle \frac{(x-5)(x-5)}{(x-5)(x+5)}=\frac{x-5}{x+5}\)

First, set up the division as the following:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}\)

Look at the leading term \(\displaystyle x^2\) in the divisor and \(\displaystyle 3x^3\) in the dividend. Divide \(\displaystyle 3x^3\) by \(\displaystyle x^2\) gives \(\displaystyle 3x\); therefore, put \(\displaystyle 3x\) on the top:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}\)

Then take that \(\displaystyle 3x\) and multiply it by the divisor, \(\displaystyle x^2-1\), to get \(\displaystyle 3x^3-3x\).  Place that \(\displaystyle 3x^3-3x\) under the division sign:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \end{array}\)

Subtract the dividend by that same \(\displaystyle 3x^3-3x\) and place the result at the bottom. The new result is \(\displaystyle 5x^2+2x+8\), which is the new dividend.

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}\)

Now, \(\displaystyle 5x^2\) is the new leading term of the dividend.  Dividing \(\displaystyle 5x^2\) by \(\displaystyle x^2\) gives 5.  Therefore, put 5 on top:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}\)

Multiply that 5 by the divisor and place the result, \(\displaystyle 5x^2-5\), at the bottom:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \end{array}\)

Perform the usual subtraction:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \\ \cline{5-7} & && & & 2x & 13 \\ \end{array}\)

Therefore the answer is \(\displaystyle 3x+5\) with a remainder of \(\displaystyle 2x+13\), or \(\displaystyle 3x+5+\frac{2x+13}{x^2-1}\).

When dividing polynomials, subtract the exponent of the variable in the numberator by the exponent of the same variable in the denominator.

If the power is negative, move the variable to the denominator instead.

First move the negative power in the numerator to the denominator:

\(\displaystyle \frac{x^{2}y^{15}}{x^{5}y^{4}z^{2}}\)

Then subtract the powers of the like variables:

\(\displaystyle \frac{y^{11}}{x^{3}z^{2}}\)