# Four Ways to Solve the Hardest SAT Math Problem on Practice Test #1

Factor Theorem – Find Two Unknown Coefficients, Given Two Factors – Polynomial – Exam Style Question
Factor Theorem – Find Two Unknown Coefficients, Given Two Factors – Polynomial – Exam Style Question

Four Ways to Solve the Hardest SAT Math Problem on Practice Test #1

Solution for New SAT Practice Test #1, Calculator-Based Math, Problem 29 (page 53)

For a polynomial p(x), the value of p(3) is −2.

Which of the following must be true about p(x)?

A) x − 5 is a factor of p(x).

B) x − 2 is a factor of p(x).

C) x + 2 is a factor of p(x).

D) The remainder when p(x) is divided by x − 3 is −2.

You can tackle this intimidating-looking problem four different ways, only one of which is as complicated as the problem seems.

Method #1: Prove a mathematical theorem while taking your test. (NOT recommended)

The College Board’s explanation (page 41) relies on Method #1. It works pretty well for Esther Godel Bach, that math genius who doesn’t prep for the SAT. Esther walks into the test not knowing the textbook method for solving the problem. She proceeds to re-create the proof for the Polynomial Remainder Theorem from scratch.

Finished with her test, Miss Bach jets off to mail in her Mensa application.

While I understand the value of proving mathematical theorems, normal people like you and I shouldn’t have to improvise proofs on a timed test of basic math concepts.

Method #2: Test the answer choices by choosing a polynomial.

I see the crazy look you’re giving me. Just how do you test answer choices when the problem doesn’t even give you a polynomial to work with?

The key lies in Question 29’s wording. “Which of the following must be true?”

If something must be true, it has to be true for each and every example.

If a bird is a swan, it must be white.

A single black swan would disprove the sentence. If the sentence is true, it has to apply to every swan in existence.

Therefore, if Question 29 is asking what must be true about p(x), we can just make up a polynomial and test the answer choices!

The question does tell us that p(3) = -2, but it doesn’t give us any other requirements. I’m going to create the simplest p(x) I can think of that makes p(3) = -2:

p(x) = -2

This may look like cheating, but notice that when x = 3, p(x) = -2, so it meets the requirements of the problem. The question uses the word must, so it doesn’t matter how simple our polynomial is as long as p(3) = -2.

Now we can test the answer choices.

Choice (A) asks whether x – 5 is a factor of p(x). When you divide a number by its factor, you get a remainder of zero. For example, 25/5 has no remainder because 5 is a factor of 25, but 29/5 has a remainder of 4 because 5 isn’t a factor of 29.

Let’s try dividing our very simple p(x) by x – 5.

x – 5 doesn’t go into -2, does it? x – 5 goes into -2 zero times (none at all!), and -2 is left over. The leftover number, -2, is the remainder. This answer is consistent with the one you’d get by doing polynomial long division.

The remainder isn’t zero, so we know that x – 5 is not a factor of -2. Choice (A) is therefore incorrect.

Choices (B) and (C) are wrong for the same reason. Dividing -2 by either x – 2 or x + 2 gives a non-zero remainder, so neither x – 2 nor x + 2 is a factor of -2.

Let’s test choice (D) in the same way.

x – 3 goes into -2 zero times with -2 left over, which is another way of saying that -2 is the remainder. This is exactly what choice (D) tells us should happen. (D) isn’t automatically true, though. Remember, the question is asking what must be true for p(x), so showing that (D) is true for p(x) = -2 doesn’t prove that it’s true for other possible polynomials where p(3) = -2.

Even though we’re not 100% sure that (D) is true, we’ve already eliminated choices (A), (B), and (C). (D) is the only one left. Since the test has to give us a valid answer choice, (D) has to be the correct answer.

An advanced student will remind me that I could have picked other polynomials that satisfy p(3) = -2, such as p(x) = x -5. While that’s true, it would have made the problem trickier and would also have failed to eliminate choice (A), making it necessary to test yet another polynomial to decide whether (A) or (D) is the correct answer.

Method #3: Test the answer choices by factoring.

Choice (A) asks whether x − 5 is a factor of p(x). In other words, if we factor p(x), we should get something like this:

p(x) = (x − 5)(x blah blah blah)(x blah blah blah)

From the factored form, we can see that when x = 5, the polynomial has to equal zero, since (x − 5) = 0 when x = 5. That means that if x − 5 is really a factor of p(x), then p(5) = 0.

The problem doesn’t tell us what p(5) is, so we don’t know if it’s zero. We do know, however, that the problem asks us what must be true. Since p(5) = 0 doesn’t have to be true, choice (A) is wrong.

Choices (B) and (C) are wrong for the same reason.

Since the SAT has to give us at least one valid answer, (D) is correct by the process of elimination.

Method #4: Use the Polynomial Remainder Theorem.

While this isn’t the only way to solve Problem 29, problems involving the Polynomial Remainder Theorem show up on the Math Level 2 Subject Test. If you’re already pretty good at math and want to raise your score, learning how to tackle remainders will increase your speed and accuracy.

Let’s start by reviewing an easy problem.

Q: What’s the remainder of 13 divided by 5?

A: 5 goes into 13 twice with 3 left over, so 13/5 is 2 with a remainder of 3.

This may be blindingly obvious, but let’s write out the math anyway. There’s a direct connection between easy math and the Polynomial Remainder Theorem.

A polynomial division problem looks a lot like 13/5, but it uses letters instead of numbers.

Notice that the remainder in the equation above is R, not R/(x-a), just like the remainder for 13/5 is 3, not 3/5.

Now here’s the Polynomial Remainder Theorem. Store it in your calculator.

The remainder, R, is equal to P(a). In other words, all you have to do to find the remainder of the division problem above is to set x = a and then find the value of P.

Q: To test your knowledge, find the remainder for the division problem below:

A: Dig into your calculator’s memory and pull up the Polynomial Remainder Theorem. (You did store it, didn’t you?)

Notice that

Since then remainder R = P(a) and a = 2, all you have to do is find R = P(2), which is

I did the above calculation pretty quickly on the Casio PRIZM by storing 2 into x and then re-typing the the polynomial.

The remainder is 2.

You can check the answer using polynomial long division if you want to.

Now let’s apply the Polynomial Remainder Theorem to Problem 29.

The question asks, for a polynomial p(x), what must be true if p(3) = -2.

This should look very familiar to you, as it’s just the Polynomial Remainder Theorem in disguise. Pull it up on your calculator again. (You did store the theorem, didn’t you?)

Since p(3) = -2 and R = P(a), a must be 3, and R must be -2.

In other words, when p(x) is divided by x – a, which is x – 3 in this case, the remainder is -2. This must be true, or else the Polynomial Remainder Theorem would be false.

Now look at the answer choices.

(A), (B), and (C) could possibly be true. We don’t know for sure that they’re false, at least. Since Question 29 asks what must be true, though, the answer can’t be (A), (B), or (C) unless we there’s a way to show for sure that one of them has to be true.

Choice (D) restates what we just showed using the Polynomial Remainder Theorem. Since we know that the remainder must be -2, choice (D) is the correct answer.

We’ve done a lot of work using Method #3, but most of that effort was spent learning and practicing the Polynomial Remainder Theorem. With enough practice, you can solve remainder problems in less than thirty seconds without having to do any polynomial long division.

Here’s some extra practice at Khan Academy:

Comment below if you have any questions!

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