\(\displaystyle (5x+2)^{2}\) is equivalent to \(\displaystyle (5x+2)(5x+2)\).

Using the FOIL method, you multiply the first number of each set \(\displaystyle 5x\cdot 5x=25x^{2}\), multiply the outer numbers of each set \(\displaystyle 5x\cdot 2=10x\), multiply the inner numbers of each set \(\displaystyle 2\cdot 5x=10x\), and multiply outer numbers of each set \(\displaystyle 2\cdot 2=4\).

Adding all these numbers together, you get \(\displaystyle 25x^{2}+10x+10x+4=25x^{2}+20x+4\). 

FOIL the first two terms:

\(\displaystyle (4x-9)(4x-9)=16x^2-36x-36x+81 = 16x^2-72x+81\)

Next, multiply this expression by the last term:

\(\displaystyle (16x^2-72x+81)(4x-9)=64x^3-144x^2-22x^2+648x+324x-729\)

Finally, combine the terms:

\(\displaystyle (4x-9)^3=64x^3-432x^2+972x-729\)

Plug in \(\displaystyle 2\) for \(\displaystyle q\) in the equation \(\displaystyle q(q-7)^{2}\)

That gives: \(\displaystyle 2(2-7)^{2}\)

Then solve the computation inside the parenthesis: \(\displaystyle 2(-5)^{2}\)

The answer should then be \(\displaystyle 50\)

Use FOIL and be mindful of exponent rules. Remember that when you multiply two terms with the same bases but different exponents, you will need to add the exponents together.

Remember to add exponents when two terms with like bases are being multiplied.

Use the FOIL method to simplify the following expression:

\(\displaystyle (x^3+2x^2)^2\)

Step 1: Expand the expression.

\(\displaystyle (x^3+2x^2)(x^3+2x^2)\)

Step 2: FOIL

First: \(\displaystyle x^3\cdot x^3 = x^6\)

Outside: \(\displaystyle x^3 \cdot 2x^2 =2x^5\)

Inside: \(\displaystyle 2x^2 \cdot x^3 = 2x^5\)

Last: \(\displaystyle 2x^2 \cdot 2x^2 = 4x^4\)

Step 2: Sum the products.

\(\displaystyle x^6+2x^5+2x^5+4x^4\)

\(\displaystyle x^6+4x^5+4x^4\)

Using FOIL on \(\displaystyle (a^3 + b^2c^2) \cdot (a^7 + bc^3)\), we see that:

First: \(\displaystyle a^3 \cdot a^7 = a^{10}\)

Outer: \(\displaystyle a^3 \cdot bc^3 = a^3bc^3\)

Inner: \(\displaystyle b^2c^2 \cdot a^7 = a^7b^2c^2\)

Last: \(\displaystyle b^2c^2 \cdot bc^3 = b^3c^5\)

Note that the middle terms are not additive: while they share common variables, they do not share matching exponents.

Thus, we have \(\displaystyle a^{10} + a^7b^2c^2 + a^3bc^3 + b^3c^5\). The arrangement goes by highest leading exponent, and alphabetically in the case of the last two terms.

Using FOIL, we see that \(\displaystyle (3x^3 + 7x^2)^2 = (3x^3 + 7x^2) \cdot(3x^3 + 7x^2)\)

First = \(\displaystyle 3x^3 \cdot 3x^3 = 9x^6\)

Outer = \(\displaystyle 3x^3 \cdot 7x^2 = 21x^5\)

Inner = \(\displaystyle 7x^2 \cdot 3x^3 = 21x^5\)

Last = \(\displaystyle 7x^2 \cdot 7x^2 = 49x^4\)

Remember that terms with like exponents are additive, so we can combine our middle terms:

\(\displaystyle 21x^5 + 21x^5 = 42x^5\)

Now order the expression from the highest exponent down:

\(\displaystyle 9x^6 + 42x^5 + 49x^4\)

Thus,

\(\displaystyle (3x^3 + 7x^2)^2 = 9x^6 + 42x^5 + 49x^4\)

\(\displaystyle (x^{2}y^{4}+xy^{6})^{2}\)

\(\displaystyle (x^{2}y^{4}+xy^{6})(x^{2}y^{4}+xy^{6})\)

We will need to FOIL.

First: \(\displaystyle x^2y^4*x^2y^4=x^4y^8\)

Inside: \(\displaystyle xy^6*x^2y^4=x^3y^{10}\)

Outside: \(\displaystyle x^2y^4*xy^6=x^3y^{10}\)

Last: \(\displaystyle xy^6*xy^6=x^2y^{12}\)

Sum all of the terms and simplify.

\(\displaystyle x^4y^8+x^3y^{10}+x^3y^{10}+x^2y^{12}\)

\(\displaystyle x^4y^8+2x^3y^{10}+x^2y^{12}\)

First, merely FOIL out your values.  Thus:

\(\displaystyle (xy^2 + 2x^3y^2)(xy^3+3x^2)\) becomes

\(\displaystyle (xy^2* xy^3+ xy^2*3x^2 + 2x^3y^2*xy^3 + 2x^3y^2*3x^2)\)

Now, just remember that when you multiply similar bases, you add the exponents.  Thus, simplify to:

\(\displaystyle x^2y^5+ 3x^3y^2 + 2x^4y^5 + 6x^5y^2\)

Since nothing can be combined, this is the final answer.