Use the foil method:  (3x - 3) (x + 7) = 3x² +21x - 3x - 21 = 3x² +18x -21 so t = 18.

use FOIL to factor the expression.

First: (x³)(x) = x⁴

Outside (x³)(7) = 7x³

Inside (–3)(x) = –3x  (Don't forget the negatives!)

Last (3)(7) = –21

Quantity A: 2² + 3² = 4 + 9 = 13

Quantity B: (2 + 3)² = 5² = 25, so Quantity B is greater.

We can also think of this in more general terms.  x² + y² does not generally equal (x + y)².

The difference of squares formula says (x + a)(x - a) = x² - a².

Thus, Quantity A equals 8.

Therefore, Quantity B is greater. 

To approach this problem, consider the two quantities

Quantity A: $\displaystyle (x+y)^2$

Quantity B: $\displaystyle x^2+4xy+y^2$

They are in different forms, so expand quantity A:

Quantity A: $\displaystyle x^2+2xy+y^2$

Quantity B: $\displaystyle x^2+4xy+y^2$

Now, for the purpose of comparison, subtract shared terms from each quantity:

Quantity A*: $\displaystyle 0$

Quantity B*: $\displaystyle 2xy$

Both $\displaystyle x$ and $\displaystyle y$ are negative, non-zero values. Since $\displaystyle 2xy$ is a product of two negative values, it must be positive. Quantity B must be greater than Quantity A.

Use the method of FOIL (First, Outside, Inside, Last) and add exponents for like bases:

$\displaystyle (xy^3+x^2y)(xy-x^3y^2)$

$\displaystyle xy^3(xy)-xy^3(x^3y^2)+x^2y(xy)-x^2y(x^3y^2)$

$\displaystyle x^2y^4-x^4y^5+x^3y^2-x^5y^3$

$\displaystyle -x^5y^3-x^4y^5+x^3y^2+x^2y^4$

Begin by expanding Quantity A:

$\displaystyle (x+y)^3$

$\displaystyle x^3+3x^2y+3xy^2+y^3$

Now in order to compare this to Quantity B:

$\displaystyle x^3+y^3$

A good method would be to subtract shared terms from each Quantity; in this case, both quantities have an $\displaystyle x^3$ and $\displaystyle y^3$ term. Removing them gives:

Quantity A' : $\displaystyle 3x^2y+3xy^2$

Quantity B' : $\displaystyle 0$

The question now is the sign of Quantity A'; if it's always positive, Quantity A is greater. If it's always negative, Quantity B is greater. If it is zero, the two are the same.

We only know that

$\displaystyle x< 0$

$\displaystyle y>0$

If $\displaystyle x=-1;y=1$, then Quantity A' would be zero.

If $\displaystyle x=-2;y=1$, then Quantity A' would be positive.

Since values of x and y can be chosen to vary the relationship, th relationship cannot be determined.

Begin by expanding Quantity A:

$\displaystyle (x+y)^3$

$\displaystyle x^3+3x^2y+3xy^2+y^3$

Now in order to compare this to Quantity B:

$\displaystyle x^3+y^3$

A good method would be to subtract shared terms from each Quantity; in this case, both quantities have an $\displaystyle x^3$ and $\displaystyle y^3$ term. Removing them gives:

Quantity A' : $\displaystyle 3x^2y+3xy^2$

Quantity B' : $\displaystyle 0$

The question now is the sign of Quantity A'; if it's always positive, Quantity A is greater. If it's always negative, Quantity B is greater. If it is zero, the two are the same.

We  know that

$\displaystyle x< 0$

$\displaystyle y>|x|$

Now compare $\displaystyle 3x^2y$ and $\displaystyle 3xy^2$: 

Looking at absolute values so that we're only considering positive terms:

$\displaystyle |3xy|$

$\displaystyle y>|x|$

From this it follows that by multiplying $\displaystyle |3xy|$ across the inequality :

$\displaystyle |3xy^2|>|3x^2y|$

From this we can determine that the magnitude of $\displaystyle |3xy^2|$ is greater. However, since this is the product of one negative number and two positive numbers, $\displaystyle 3xy^2$ is negative, and the sum of $\displaystyle 3xy^2$ and $\displaystyle 3x^2y$ must in turn be negative, and so Quantity A' must be negative!

From this we can say that Quantity B is greater.

This problem is deceptive. Looking at Quantity A, one may think to factor and reduce it as follows:

$\displaystyle \frac{x^2+5x-14}{x-2}$

$\displaystyle \frac{(x+7)(x-2)}{x-2}$

$\displaystyle x+7$

Which is identical to Quantity B.

However, we cannot ignore that $\displaystyle x-2$ in the original fraction! We are given no conditions as to the value of $\displaystyle x$. If $\displaystyle x=2$, then Quantity A would be undefined. Since we're not given the condition $\displaystyle x \neq 2$, we cannot ignore this possibility.

The relationship cannot be established.

You must FOIL the expression which means to multiply the first terms together followed by the outer terms, then the inner terms and lastly, the last terms.

The expression written out looks like 

$\displaystyle (x-2)(x-2)$.  

You multiple both First terms to get $\displaystyle x^{2}$.  

Then the outer terms are multiplied $\displaystyle (x*-2=-2x)$.

Then you multiple the inner terms together $\displaystyle (-2*x=-2x)$.  

Finally you multiply the last terms of each $\displaystyle (-2*-2=4)$.  

This gives you $\displaystyle x^{2}-2x-2x+4$ or $\displaystyle x^{2}-4x+4$.