To multiple these binomials, you can use the FOIL method to multiply each of the expressions individually.This will give you

\(\displaystyle (x)(3x)+(x)(2)+(-9)(3x)+(-9)(2)\)

or \(\displaystyle 3x^{2}-25x-18\).

To add these two trinomials, you will first begin by combining like terms. You have two terms with \(\displaystyle x^2\), two terms with \(\displaystyle x\), and two terms with no variable. For the two fractions with \(\displaystyle x^2\), you can immediately add because they have common denominators:

\(\displaystyle (\frac{1}{2}x^2 +2x - 4) + (\frac{3}{2}x^2 -5x -4)\)

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

\(\displaystyle 2x^2 -3x - 8\)

Factoring the expression gives \(\displaystyle \small \frac{}{}\)\(\displaystyle \small \small \frac{(x+3)(x+2)}{(x-4)(x+4)(x+2)}\). Values that are in both the numerator and denominator can be cancelled. By cancelling \(\displaystyle \small x+2\), the expression becomes \(\displaystyle \small \small \frac{(x+3)}{(x-4)(x+4)}\).

By factoring the equation you get \(\displaystyle \small \small \small \frac{(x+3)(x-5)}{(x+2)(x-5)}\). Values that are in both the numerator and denominator can be cancelled. Cancelling the \(\displaystyle \small x-5\) values gives \(\displaystyle \small \small \small \frac{x+3}{x+2}\).

By foiling the binomials, multiplying the firsts, then the outers, followed by the inners and lastly the lasts, the expression you get is:

 \(\displaystyle \small 2(x^2+4x+2x+8)\).

However, the expression can not be considered simplified in this state.

Distributing the two and adding like terms gives \(\displaystyle \small 2x^2+12x+16\).

Use the FOIL method to simplify the binomial.

\(\displaystyle (a+b)(c+d) = ac+ad+bc+bd\)

\(\displaystyle (ax)(2ax)+(ax)(3b)+(b)(2ax)+(b)(3b)\)

Simplify the terms.

\(\displaystyle 2a^2x^2+5abx+3b^2=8x^2+30x+27\)

Notice that the coefficients can be aligned to the unknown variables.  Solve for \(\displaystyle a\) and \(\displaystyle b\).

\(\displaystyle 2a^2 = 8 \rightarrow a^2 = 4 \rightarrow a=2\)

\(\displaystyle 3b^2 = 27\rightarrow b^2=9 \rightarrow b=3\)

\(\displaystyle a-b = 2-3 = -1\)

The answer is:  \(\displaystyle -1\)

Multiply each term of the first trinomial by second trinomial.

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

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

\(\displaystyle (-1)(3x^2-x+3) = -3x^2+x-3\)

Add and combine like-terms.

The answer is:  \(\displaystyle 3x^4-4x^3+x^2-2x-3\)

The expression will need to be rearranged from highest to lowest powers in order to be simplified.

\(\displaystyle \frac{8x-10+2x^2}{-16+16x} =\frac{2x^2+8x-10}{16x-16}\)

Factor a 2 in the numerator.

\(\displaystyle 2x^2+8x-10 = 2(x^2+4x-5)\)

Factor the term in parentheses.

\(\displaystyle 2(x^2+4x-5) = 2(x-1)(x+5)\)

Factor the denominator.

\(\displaystyle 16x-16 = 16(x-1)\)

Divide the numerator with the denominator.

\(\displaystyle \frac{2(x-1)(x+5)}{16(x-1)}\)

The expression becomes:

 \(\displaystyle \frac{x+5}{8}\)

The answer is:  \(\displaystyle \frac{1}{8}x+\frac{5}{8}\)

The correct answer is \(\displaystyle x=-6\) or \(\displaystyle x=4\). The first step of the problem is to cross multiply. This will give the following equation:

\(\displaystyle 24=x^2+2x\)

After subtracting \(\displaystyle 24\) from each side the equation looks like:

\(\displaystyle 0=x^2+2x-24\)

The expression on the right hand side can be factored into: 

\(\displaystyle 0=(x-4)(x+6 )\)

Both \(\displaystyle 4\) and \(\displaystyle 6\) satisfy the above equation and are therefore the correct answers. 

Use the FOIL method (First, Outer, Inner, Last):

\(\displaystyle x * 2x = 2x^2\)

\(\displaystyle x * 4 = 4x\)

\(\displaystyle -4 * 2x = -8x\)

\(\displaystyle -4 * 4 = -16\)

Put all of these terms together:

\(\displaystyle 2x^2 + 4x - 8x - 16\)

Combine like terms:

\(\displaystyle 2x^2 - 4x - 16\)