Via the FOIL method, we can attest that x(2x) + x(3) + –4(2x) + –4(3) = 2x² – 5x – 12.

Use FOIL: 

  (x+3)(x-5)(x) = (x² - 5x + 3x - 15)(x) = x³ - 5x² + 3x² - 15x = x³ - 2x² - 15x for A.

  (x-3)(x-1)(x+3) = (x-3)(x+3)(x-1) = (x² + 3x - 3x - 9)(x-1) = (x² - 9)(x-1)

  (x² - 9)(x-1) = x³ - x² - 9x _⁺ 9 for B. 

The difference between A and B: 

 (x³ - 2x² - 15x) - (x³ - x² - 9x _⁺ 9) = x³ - 2x² - 15x - x³ + x² + 9x - 9

 = - x² - 4x - 9. Since all of the terms are negative and x > 0:

  A - B < 0.

Rearrange A - B < 0:

  A < B

\(\displaystyle x^3+5x^2-10=2x^2\)

First, move all terms to one side of the equation to set them equal to zero.

\(\displaystyle x^3+5x^2-2x^2-10x=0\)

\(\displaystyle x^3+3x^2-10x=0\)

All terms contain an \(\displaystyle \small x\), so we can factor it out of the equation.

\(\displaystyle x(x^2+3x-10)=0\)

Now, we can factor the quadratic in parenthesis. We need two numbers that add to \(\displaystyle \small 3\) and multiply to \(\displaystyle \small -10\).

\(\displaystyle -2*5=-10\ \text{and}\ -2+5=3\)

\(\displaystyle x(x-2)(x+5)=0\)

We now have three terms that multiply to equal zero. One of these terms must equal zero in order for the product to be zero.

\(\displaystyle \begin{matrix} x=0 & x-2=0 &x+5=0 \\ x=0 & x=2 & x=-5 \end{matrix}\)

Our answer will be \(\displaystyle \small x=0,2,-5\).

In order to simplify this expression, you need to use the FOIL method. First rewrite the expression to look like this: \(\displaystyle (2x+3)(2x+3)\)

Next, multiply your first terms together: \(\displaystyle 2x\times2x =4x^{2}\)

Then, multiply your outside terms together: \(\displaystyle 2x\times3=6x\)

Then, multiply your inside terms together: \(\displaystyle 3\times2x=6x\)

Lastly, multiply your last terms together: \(\displaystyle 3\times3=9\)

Together, you have \(\displaystyle 4x^{2}+6x+6x+9\)

You can combine your like terms, \(\displaystyle 6x+6x\), to give you the final answer: \(\displaystyle 4x^{2}+12x+9\)

Use the FOIL method (first, outside, inside, last) to find the product of:

\(\displaystyle (x+9)(x-7)\)

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

Outside: \(\displaystyle -7\cdot x=-7x\)

Inside: \(\displaystyle 9\cdot x=9x\)

Last: \(\displaystyle -7\cdot 9=-63\)

Sum the products to find the polynomial:

\(\displaystyle x^2-7x+9x-63\)

\(\displaystyle x^2+2x-63\)

To solve this problem, use the FOIL method. Start by multiplying the First term in each set of parentheses: \(\displaystyle x^{2} \cdot x =x^{3}\) 

Then multiply the outside terms: \(\displaystyle x^{2}\cdot (-1) =-x^{2}\)

Next, multiply the inside terms: \(\displaystyle x \cdot x= x^{2}\)

Finally, multiply the last terms: \(\displaystyle x\cdot (-1) = -x\)

When you put the pieces together, you have \(\displaystyle 2(x^{3} -x^{2} +x^{2}-x)\). Notice that the middle terms cancel each other out, and you are left with \(\displaystyle 2(x^{3} -x)\). When you distribute the two, you reach the answer: \(\displaystyle 2x^{3} -2x\)