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 \(\displaystyle 3x^{2}+10x-8\)