Even though the expression uses letters in the place where it is common to find numbers, we should recognize it is still the multiplication of two binomials, and the FOIL process can be used here.

F: $\displaystyle x\cdot x=x^{2}$

O: $\displaystyle x\cdot -b=-bx$

I: $\displaystyle a\cdot x=ax$

L: $\displaystyle a\cdot -b=-ab$

So we have $\displaystyle x^{2}-bx+ax-ab$

You can use the FOIL method to expand the expression 

F:First 

O: Outer

I: Inner

L:Last

L-Last

$\displaystyle (x+6)(x-4)$

F: $\displaystyle x\cdot x=x^{2}$

O: $\displaystyle x\cdot -4=-4x$

I: $\displaystyle 6\cdot x=6x$

L: $\displaystyle 6\cdot -4=-24$

$\displaystyle x^{2} - 4x + 6x - 24 = x^{}2-2x-24$

Remember, FOIL stands for First-Outer-Inner-Last

Multiply the first terms 

$\displaystyle 2x\cdot3x=6x^2$

Multiply the outer terms

$\displaystyle 2x\cdot-4=-8x$

Multiply the inner terms

$\displaystyle 1\cdot3x=3x$

Multiply the last terms

$\displaystyle 1\cdot-4=-4$

Now we simply add them all together

$\displaystyle 6x^2-8x+3x-4$

And combine like-terms

$\displaystyle 6x^2-5x-4$

We distribute each term in each parentheses to the terms of the other parentheses.

We get:

$\displaystyle x*x^2, x*-7, x*x, 1*-7$

Which Simplifies:

$\displaystyle x^3, x^2,-7x,-7$

We will arrange these from highest to lowest power, and adding a sign in between terms based on the coefficient of each term:

So, the answer is $\displaystyle x^3+x^2-7x-7$

This is a quadratic equation, but it is not in standard form.

$\displaystyle (x - 5) (x -3) = 24$

We express it in standard form as follows, using the FOIL technique:

$\displaystyle x \cdot x - x \cdot 3 - 5 \cdot x + 5 \cdot 3 = 24$

$\displaystyle x ^{2}-3 x - 5x+ 15 = 24$

$\displaystyle x ^{2}-8 x + 15 = 24$

$\displaystyle x ^{2}-8 x + 15- 24 = 24 - 24$

$\displaystyle x ^{2}-8 x -9 = 0$

Now factor the quadratic expression on the left. It can be factored as

$\displaystyle (x+A)(x+B)$

where $\displaystyle AB = -9, A+B = -8$.

By trial and error we find that $\displaystyle A = -9, B = 1$, so 

$\displaystyle x ^{2}-8 x -9 = 0$ 

can be rewritten as

$\displaystyle (x -9 )(x+1) = 0$.

Set each linear binomial equal to 0 and solve separately:

$\displaystyle x - 9 = 0 \Rightarrow x = 9$

$\displaystyle x + 1 \Rightarrow x = -1$

The solution set is $\displaystyle \left \{ -1, 9\right \}$.

$\displaystyle \left (3.1x^{2}+ 4.5x + 0. 5 \right ) - (1.7x^{2} - 3.9x - 3.2)$

can be determined by subtracting the coefficients of like terms. We can do this vertically as follows:

$\displaystyle \begin{matrix} \; \; 3.1x^{2}+ 4.5x + 0. 5 \\ - \underline{(1.7x^{2} - 3.9x - 3.2) } \end{matrix}$

By switching the symbols in the second expression we can transform this to an addition problem, and add coefficients:

$\displaystyle \begin{matrix} \; \; 3.1x^{2}+ 4.5x + 0. 5 \\ \underline{-1.7x^{2} +3.9x +3.2 }\\ \; \; 1.4x^{2} + 8.4x +3.7\end{matrix}$

$\displaystyle \left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

can be determined by adding the coefficients of like terms. We can do this vertically as follows:

$\displaystyle \begin{matrix} \; \; 3.1x^{2}+ 4.5x + 0. 5 \\ + \underline{1.7x^{2} - 3.9x - 3.2 }\\ \; \; 4.8x^{2} + 0.6x -2.7\end{matrix}$

Use the difference of squares pattern

$\displaystyle (A + B)(A - B) = A^{2} - B^{2}$

with $\displaystyle A = \frac{1}{5x}$ and $\displaystyle B =5x$ :

$\displaystyle \left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

$\displaystyle =\left ( \frac{1}{5x} \right )^{2} - \left ( 5x\right )^{2}$

$\displaystyle = \frac{1}{25x^{2}} - 25x ^{2}$

$\displaystyle = \frac{1}{25x^{2}} - \frac{25x ^{2} \cdot 25x ^{2} }{25x ^{2} }$

$\displaystyle = \frac{1}{25x^{2}} - \frac{625x ^{4} }{25x ^{2} }$

$\displaystyle = \frac{1-625x ^{4} }{25x^{2}}$

$\displaystyle = \frac{-625x ^{4} + 1 }{25x^{2}}$

Use the difference of squares pattern

$\displaystyle (A + B)(A - B) = A^{2} - B^{2}$

with $\displaystyle A = \frac{1}{3x}$ and $\displaystyle B = \frac{2}{3}$ :

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

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

$\displaystyle = \frac{1}{9x^{2}} - \frac{4}{9}$

$\displaystyle = \frac{1}{9x^{2}} - \frac{4x^{2}}{9x^{2}}$

$\displaystyle = \frac{1-4x^{2}}{9x^{2}}$

$\displaystyle = \frac{-4x^{2}+ 1}{9x^{2}}$

Start by factoring the numerator. Notice that each term in the numerator has an $\displaystyle x$, so we can write the following:

$\displaystyle x^3+7x^2+10x=x(x^2+7x+10)$

Next, factor the terms in the parentheses. You will want two numbers that multiply to $\displaystyle 10$ and add to $\displaystyle 7$.

$\displaystyle x(x^2+7x+10)=x(x+5)(x+2)$

Next, factor the denominator. For the denominator, we will want two numbers that multiply to $\displaystyle 2$ and add to $\displaystyle 3$.

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

Now that both the denominator and numerator have been factored, rewrite the fraction in its factored form.

$\displaystyle \frac{x^3+7x^2+10x}{x^2+3x+2}=\frac{x(x+5)(x+2)}{(x+2)(x+1)}$

Cancel out any terms that appear in both the numerator and denominator.

$\displaystyle \frac{x(x+5)(x+2)}{(x+2)(x+1)}=\frac{x(x+5)}{x+1}$