While this problem is certainly doable with pure algebra, many complicated polynomial questions like this one are much more quickly solved by picking numbers. There are lots ofterms in this problem, but the problem specifically states that the equation must hold true for ALL values ofso you can turn a lot of complicated algebra into some rather quick arithmetic by picking an easy value for。你不能选择because the only appearance ofis when it's multiplied by, but you can pickleaving you with very small numbers to calculate. Doing so will leave you with:

And that simplifies quickly, to:

Which becomes:

And that means thatmust equal。

You could of course, do the algebra here, but as a general rule when the polynomials involve several multiplicative terms like this on the SAT, the question usually gives you an opportunity to pick a simple number to make the work much easier.

When polynomials have this many terms on the SAT, and when you see a phrase like "for all values of," it is usually faster and easier to pick a number forto test which answer choices match the given information (that's typically when the question asks "which of the following expressions is equivalent to the above?") or, as in this case, to remove the commonly-occurring variable and solve for the less-common one. Here you have lots ofterms but only oneterm, so since they say that this equation holds true for ALL VALUES of, you can pick an easy-to-use value likeand then use arithmetic, not multi-step algebra, to solve for。If you plug inthe given equation becomes:

Which quickly simplifies to:

And that simplifies to:

So, meaning that。

Note that you could of course do the algebra, distributing all the multiplication in the given equation and then combining like terms. But on most polynomial questions with this much multiplication to do, the shortcut of picking a small number can save you lots of time and be less error-prone.

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 ofwill produce the same output for the correct answer.

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

Then plug in that sameto the answer choices looking for matches. You'll find thatgives you that match:

leads togiving 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 likewithout doing the work. Why? You need the numeric terms in that multiplication to multiply toand 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, theterm:

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 exactlyas the numeric term. That leaves only one choice remaining,。That's the only choice that gives you a cubed term ofand 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 forand 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 towhen you useis 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, usehere. 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 likeand 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 forand perform the arithmetic than it is to work through the algebra. Because this equation is true for "all values of" (the prohibitionis there to ensure that the denominator does not equal zero), you can choose any value forand 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 usingbecause it's such an easy number to cube, square, and multiply by. If you do so, the given expression becomes:

When you then pluginto 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 foranyvalue 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!)