To FOIL this binomial distribution, we simply distribute the terms in a specific order:

\(\displaystyle (7x+5)(2x+5)\)

Multiply the First terms:

\(\displaystyle 7x\cdot 2x = 14x^2\) 

Multiply the Outer terms:

\(\displaystyle 7x\cdot 5 = 35x\)

Multiply the Inner terms:

\(\displaystyle 5\cdot 2x = 10x\)

Multiply the Last terms:

\(\displaystyle 5\cdot 5 = 25\)

Lastly, combine any terms that allow this (usually, but not always, the two middle terms):

\(\displaystyle 35x+10x = 45x\)

Arrange your answer in descending exponential form, and you're done.

\(\displaystyle (7x+5)(2x+5) = 14x^2 + 45x +25\)

To FOIL this binomial distribution, we simply distribute the terms in a specific order:

\(\displaystyle (ax+by)(ax+by)\)

Multiply the First terms:

\(\displaystyle ax\cdot ax = a^2x^2\) 

Multiply the Outer terms:

\(\displaystyle ax\cdot by = abxy\)

Multiply the Inner terms:

\(\displaystyle by\cdot ax = abxy\)

Multiply the Last terms:

\(\displaystyle by\cdot by = b^2y^2\)

Lastly, combine any terms that allow this (usually, but not always, the two middle terms):

\(\displaystyle abxy+abxy = 2abxy\)

Arrange your answer in descending exponential form, and you're done.

\(\displaystyle (ax+by)(ax+by) = a^2x^2 + b^2y^2+2abxy\)

To FOIL this binomial distribution, we simply distribute the terms in a specific order:

\(\displaystyle (x-4)(x+7)\)

Multiply the First terms:

\(\displaystyle x\cdot x = x^2\) 

Multiply the Outer terms:

\(\displaystyle x\cdot 7 = 7x\)

Multiply the Inner terms:

\(\displaystyle -4\cdot x = -4x\)

Multiply the Last terms:

\(\displaystyle -4\cdot 7 = -28\)

Lastly, combine any terms that allow this (usually, but not always, the two middle terms):

\(\displaystyle 7x-4x = 3x\)

Arrange your answer in descending exponential form, and you're done.

\(\displaystyle (x-4)(x+7) = x^2+3x-28\)

The trick to this expression is to remember that only those terms which share both common variables AND common exponents are additive. In other words, you cannot add \(\displaystyle a^3 + b^3\) any more than you can add \(\displaystyle b^3 + b^{300}\).

To FOIL this binomial distribution, we simply distribute the terms in a specific order:

\(\displaystyle (2a^3+ 4x^2)(-b^3 - 5y^2)\)

Multiply the First terms:

\(\displaystyle 2a^3\cdot -b^3 = -2a^3b^3\) 

Multiply the Outer terms:

\(\displaystyle 2a^3\cdot -5y^2 = -10a^3y^2\)

Multiply the Inner terms:

\(\displaystyle 4x^2\cdot -b^3 = -4b^3x^2\)

Multiply the Last terms:

\(\displaystyle 4x^2\cdot -5y^2 = -20x^2y^2\)

Lastly, combine any terms that allow this (usually, but not always, the two middle terms). In this case, no two terms are compatible.

Arrange your answer in descending exponential form, and you're done.

\(\displaystyle (2a^3+ 4x^2)(-b^3 - 5y^2) = -2a^3b^3 - 10a^3y^2 - 4b^3x^2 - 20x^2y^2\)

Use the grid method to FOIL.

Combine the like terms.

Using FOIL which stands for the multiplication process between the Firsts, Outers, Inners, and Lasts, we end up with the expression 

\(\displaystyle 3a^6-15a^3+6a^3-30\).

From there, combine the like terms \(\displaystyle -15a^3+6a^3\) to get \(\displaystyle -9a^3\).

Therefore the product becomes,

\(\displaystyle 3a^6-9a^3-30\)

This question is asking you to multiply two binomials. You can use the grid method for FOIL.

Using our properties of exponents, we could rewrite \(\displaystyle f(x)\) as \(\displaystyle \sqrt[5]{(x-2)^{4}}+3\) 

This means that when we input \(\displaystyle x\), we first subtract 2, then take this to the fourth power, then take the fifth root, and then add three. We want to look at these steps individually and see whether there are any values that wouldn’t work at each step. In other words, we want to know which \(\displaystyle x\) values we can put into our function at each step without encountering any problems.

The first step is to subtract 2 from \(\displaystyle x\). The second step is to take that result and raise it  to the fourth power. We can subtract two from any number, and we can take any number to the fourth power, which means that these steps don't put any restrictions on \(\displaystyle x\).

Then we must take the fifth root of a value. The trick to this problem is recognizing that we can take the fifth root of any number, positive or negative, because the function \(\displaystyle x^{\frac{1}{5}}\) is defined for any value of \(\displaystyle x\); thus the fact that \(\displaystyle f(x)\) has a fifth root in it doesn't put any restrictions on \(\displaystyle x\), because we can add three to any number; therefore, the domain for \(\displaystyle f(x)\) is all real values of \(\displaystyle x\).

The domain of the function is all real numbers except x = –3. When x = –3, f(–3) is undefined.

When x = 0 or x = 3, the function is undefined due to its denominator. 

Thus the domain is all real numbers x, such that x is not equal to 0 or 3.