On a question like this, where your job is to match a given polynomial with one of the polynomials in the answer choices, you'll find that it can be difficult to artfully factor the given polynomial to match the factored answer choices. So your choices are really to 1) expand the multiplication in each answer choice until you find a match, or 2) pick a small number for the variable and see which answer choices produce the same result. After all, if the expressions are equivalent, then the same input value of  will produce the same output for the correct answer.

Here you might choose  to make for easier arithmetic. That would mean that the given expression is:

Then plug in that same  to the answer choices looking for matches. You'll find that  gives you that match:

 leads to  giving you your only match.  And note that while you should try the remaining answer choices even once you've found a match, you can eliminate choices like  without doing the work. Why? You need the numeric terms in that multiplication to multiply to  and here a quick test, just multiplying the numeric, non-variable terms here, gives you , so you can prove without doing much math that it's not a candidate.

When you face polynomial questions with more than 3 or 4 terms on the SAT, there is usually a shortcut available to you. Here you could attempt distribute all the multiplication in the given expression to see which answer it matches; as this solution does so, however, notice the shortcuts that these steps open up for you.

To distribute the multiplication, make sure you're very organized. Here you may wish to start by distributing the first term, , across both terms in the second parentheses. That will give you:

Now notice that new cubed term, . Nowhere else in this multiplication will you multiply terms that will give you a third power, so you can scan the answer choices and eliminate any terms that don't have a cubed term of . That can help you eliminate .

Next you may multiply the second term, , in the three-term parentheses, distributing across the second parentheses set:

And then do the same for the third, the  term:

You should see here that there is only one numerical term in all this expanded multiplication, and it's . So again you can scan the answer choices getting rid of any choices that won't lead to exactly  as the numeric term. That leaves only one choice remaining, . That's the only choice that gives you a cubed term of  and a numeric term of . Of course, you may wish to finish the job with the multiplication you've done.  You have as your terms:

This simplifies to:

And since each term is divisible by two, you can factor to get to exactly the right answer, 

This problem asks you to distribute multiplication across two parenthetical polynomials, then combine like terms to arrive at an equivalent expression. You can gain some insight into the algebra that must be done by looking at the answer choices: each choice is a list of terms added/subtracted with no parentheses, so that tells you need to apply the multiplication in the given expression to remove those parentheses.

When doing so in such a large polynomial, you must be organized and careful. Unlike with a 2x2 set of parentheses, you cannot simply apply FOIL here but the same logic applies: you need to multiply each term in the first parentheses by each term in the second. So it helps to do that multiplication one term at a time:

First term by second parentheses: 

Second term by second parentheses: 

Third term by second parentheses: 

So your list of terms is now: 

And then you can combine like terms to arrive at the answer: 

Note also that on these problems that ask for equivalent expressions, you also have the option to plug in a small number for  and then test which answer choices arrive at the same value. Here if you say that , the given expression becomes .  The only answer choice that also amounts to  when you use  is the correct answer:  becomes .

When you're asked to work with multi-term polynomials that involve distributed multiplication, a nice shortcut appears when you see a phrase such as "for all values of ." Since the question is saying that the equation is true for any value of , you have the option to just pick an easy-to-work-with value and use numbers instead of variables. You might, for example, use  here.  If you do so, then the given equation looks like:

And since 1 is such an easy number to multiply, square, and cube, this simplifies quite quickly to:

This then simplifies to:

So , meaning that .

Whenever a large polynomial equation is accompanied by the words "for all values of..." you have the option to simply pick a number for that variable rather than perform the cumbersome algebra. Since the equation here must hold true for any value of , you can pick an easy-to-work-with value like  and use that to solve for . If you do so, you'll have:

But since 1 is such an easy number to square, cube, or multiply by, your scratchwork probably looks more like:

That simplifies to:

So 

Often times when polynomial problems involve this many terms, it is much faster to choose a value for  and perform the arithmetic than it is to work through the algebra. Because this equation is true for "all values of " (the prohibition  is there to ensure that the denominator does not equal zero), you can choose any value for  and dramatically shortcut your work. If you were to choose, for example, , the equation becomes:

Combining like terms simplifies this to:

This then becomes:

So 

When a question asks you to find equivalent expressions, you have two options: 1) you can perform algebra to make the answer choices look like the question (or the question look like the answer choices), or 2) you can pick small, easy to work with numbers and see which answer choice gives you the same output as the given expression. That's because if the expressions are truly equivalent, then they will produce the same answer for the same value of .

Here you might consider using  because it's such an easy number to cube, square, and multiply by.  If you do so, the given expression becomes:

When you then plug  into each of the answer choices, you're looking for an output of . Only one answer choice does that:

 becomes:

, which simplifies to:

Though we could solve this question algebraically, the question stem has given us a very powerful hint! "For all values of x" means that the equation should hold true for any value of x. So - let's pick a value!

If we choose x = 1, we can drastically simplify the stem in one step, as 

(3x^2 + 4x - 2)(xy) = x^3+2x^2-4 

becomes

(3+4-2)(y) = 1+2-4

From here, if we combine like terms to arrive at 

5y = -1

and by dividing both sides by 5 to isolate y, we find that 

(*note - be suspicious of the answer choice "it cannot be determined." This answer option is almost never a gift, and generally needs to be proven if correct. Chances are, there's something you can do with the information if you dig in and search for clues in the question stem and answer options!)

The quadratic equation  factors to . Remember, to solve for the solutions of a quadratic you can factor it to two parenthetical terms multiplied together, because then you can leverage the fact that anything times zero is zero. If either of those parentheticals were to equal zero, then the entire equation equals zero.

The most common place to make a mistake on quadratic problems is to properly factor as above but to not take the final step of setting each parenthetical term equal to zero. Here to solve you need to perform:

, so  is the solution to the first parentheses.

, so  is the solution to the second parentheses.

Any time you see  and  terms in an equation, you're likely dealing with a quadratic or polynomial, and it is therefore likely that the best way to solve those is to move all the terms to one side of the equation to set it equal to zero. Here you can do that by subtracting  from both sides to arrive at the quadratic:

Now your job is to factor the quadratic, looking for two values that multiply to  and sum to .  This should lead you to a factored quadratic of:

The final - and often missed when working under time pressure! - step is to set each parentheses equal to zero to solve the equation.  This means that your solutions are:

, so 

, so