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 for to 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 of terms but only one term, so since they say that this equation holds true for ALL VALUES of , you can pick an easy-to-use value like and then use arithmetic, not multi-step algebra, to solve for . If you plug in the 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.

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!)

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

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 .

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.

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

This quadratic offers a great shortcut for those fluent in the art of factoring quadratics. You know that is one of the solutions, meaning that the quadratic must factor to:

Where you now just need to determine how to fill the second parentheses. And you know that when factoring a quadratic, you need to multiply to the last term and sum to the middle term. And here you know the middle term has a coefficient of . So in order to arrive at a sum of , the factorization must be:

, meaning that the other solution is .

Of course, there's a "long way" on this problem that isn't that much longer. If you know that is a solution, you can plug in to the given quadratic and solve for :

So

Then you can plug that back into the original and factor the quadratic:

So the solutions are , which you were given, and , the answer you need.

To simplify, remove parentheses and combine like terms:

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.