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:

$\displaystyle 2x^2 - 4x - 16$