This is a FOIL-ing problem. First, set up the numbers in a form we can use to create the function.

Take the opposite sign of each of the numbers and place them in this format. $\displaystyle (x-5)(x+8)$

Multiply the $\displaystyle x$ in the first parentheses by the $\displaystyle x$ and 8 in the second parentheses respectively to get $\displaystyle x^{2}+8x$

Multiply the $\displaystyle -5$ in the first parentheses by the $\displaystyle x$ and 8 in the second parentheses as well to give us $\displaystyle -5x-40$.

Then add them together to get $\displaystyle x^{2}+8x-5x-40$

Combine like terms to find the answer which is $\displaystyle x^{2}+3x-40$.

Simplify using FOIL method.

Remember that multiplying variables means adding their exponents.

F: $\displaystyle 3x^{2}*2x^{3} = 6x^{5}$

O: $\displaystyle 3x^{2}*(-8) = -24x^{2}$

I: $\displaystyle -3 *2x^{3}=-6x^{3}$

L: $\displaystyle -3 *-8= 24$

Combine the terms. Note that we cannot simplify further, as the exponents do not match and cannot be combined.

$\displaystyle 6x^{5}-6x^{3}-24x^{2}+24$

$\displaystyle \textup{FOIL: Product is the sum of the products of the first, outer, inner and last terms:}$

$\displaystyle \left ( 3x+2\right )\left ( 2x-5\right )=$ $\displaystyle 3x\ast2x+3x\left ( -5\right )+ 2\ast2x+2\left ( -5\right )$

$\displaystyle 6x^{2}-15x+4x-10\;\;\;=\;\;\;6x^{2}-11x-10$

To find the factors, you must determine which of the sets of factors result in the polynomial when multiplied together. Using the FOIL method, a set of factors with the form $\displaystyle (x+a)(x+b)$ will result in $\displaystyle x^{2}+ax+bx+ab$. Applying this format to the given equation of $\displaystyle x^{2}+11x+24$, $\displaystyle a+b$ must equal 11 and $\displaystyle ab$ must equal 24. The only set that works is $\displaystyle (x+8)(x+3)$.

If you use the FOIL method, you will multiply each expression individually. So, $\displaystyle (x+3)(2x-5)$ becomes $\displaystyle (x*2x)+(x*-5)+(3*2x)+(3*-5)$, which simplifies to $\displaystyle 2x^{2}+x-15$.

$\displaystyle x^{2}y(2x + 3y)$

Use the distributive property to simplify the expression.  In general, $\displaystyle a(b+c) = ab + ac$.

$\displaystyle x^{2}y(2x + 3y)=(x^{2}y*2x) + (x^{2}y*3y)$

Now we can begin to combine like terms through multiplication.

$\displaystyle (x^{2}y*2x) + (x^{2}y*3y)$

$\displaystyle (2x^{3}y)+(3x^2y^2)$

We cannot simplify further.

The FOIL method yields the products below.

First: $\displaystyle 4x* 2x=8x^{2}$

Outside: $\displaystyle 4x* 6=24x$

Inside: $\displaystyle -7* 2x=-14x$

Last: $\displaystyle -7* 6=-42$

Add these four terms, and combine like terms, to obtain the product of the binomials.

$\displaystyle 8x^{2}+24x+(-14x)+(-42)=8x^{2}+10x-42$

$\displaystyle x^3-3x^2-18x$

First, factor out an $\displaystyle x$, since it is present in all terms.

$\displaystyle x(x^2-3x-18)$

We need two factors that multiply to $\displaystyle -18$ and add to $\displaystyle -3$.

$\displaystyle -6*3=-18$ and $\displaystyle -6+3=-3$

Our factors are $\displaystyle -6$ and $\displaystyle +3$.

$\displaystyle x(x-6)(x+3)$

We can check our answer using FOIL to get back to the original expression.

First: $\displaystyle (x)(x)=x^2$

Outside: $\displaystyle (x)(3)=3x$

Inside: $\displaystyle (x)(-6)=-6x$

Last: $\displaystyle (-6)(3)=18$

Add together and combine like terms.

$\displaystyle x^2+3x-6x-18=x^2-3x-18$

Distribute the $\displaystyle x$ that was factored out first.

$\displaystyle x(x^2-3x-18)=x^3-3x^2-18x$

$\displaystyle \textup{FOIL to distribute:} \left ( 2p+4\right )\left ( p-5\right )= 2p(p)+2p(-5)+4(p)+4(-5)$

$\displaystyle 2p^{2}-10p+4p-20\;\;\;\;\;\;\;\;2p^{2}-6p-20$

To expand $\displaystyle (x+4)(3x-2)$, use the FOIL method, where you multiply each expression individually and take their sum. This will give you

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

or $\displaystyle 3x^{2}+10x-8$