Posts tagged with: factoring

f(x) = x^3 – cx^2 + 4x – 4c

In the function f above, c is a constant. How many x-intercepts does the function have?

Can you show how to solve this through logic/algebra? TIA!!

All about factoring here. The giveaway pattern that jumped out to me was the c‘s. Let’s see what we can factor out of first two terms and then what we can factor out of the second two terms.


Once you’ve got that factored, it’s a little easier to see how many x-intercepts there should be. You’ll have an x-intercept for every time you can make the value of the expression zero for some value of x. But can x^2+4 ever equal zero? Nope! So really, the only time the expression will ever equal zero is when x=c. Therefore, there is only one x-intercept for that function.


test number 4 section 4 question number 25 indicates
Choice B is correct.
they the college board explains:
In f(x), factoring out the greatest common factor, 2x, yields
f(x) = 2x (x2 + 3x + 2).

however I do not find 2x to be a factor of X^3 +6X +4 because the 4 has no X associated with it. I did not get the answer as b either but c. Could you explain?

Thx John

I’ve got a full explanation of that question here (you can see all the official test questions I’ve explained here).

I think the problem you’re having is related to the initial factoring. You’re originally told that f(x)=2x^3+6x^2+4x. To factor a 2x out of that, you need to pull it out of each term (i.e, out of each part of the expression separated by a plus or minus sign). Here’s what your work might look like:


Does that help?

Hi Mike, I was wondering if you could explain a little bit better question 2 of the “Binomial Squares and Difference of Two squares” Practice questions. More particularly, I was wondering if there was a way to convert the equation to the right answer.

You mean factoring the given expression rather than expanding the answer choices? Yes, that can be done. Here’s the original expression:


Note the patterns in the powers of a and b: you have a^6, a^3, and no a in the final term; you have no b in the first term, then b and b^2. With enough experience looking at binomials, you might recognize that as a pattern that you’d expect to get when you FOILed something like this:


From there, recognize that you have perfect squares as coefficients in the first and third terms: 25 and 16. That’s pretty cool—let’s see if using 5 and 4 in the factored expression gives us what we want:


Yep, that’s what we were hoping for, so our factoring was correct. That means choice A is correct.

The problem with which I’m having some difficulty, taken from the chapter on the Passport to Advanced Math section in The Official SAT Study Guide book (page 229), is as follows:

(y^5) – (2y^4) – (cxy) + (6x)

In the polynomial above, “c” is a constant. If the polynomial is divisible by “y – 2,” what is the value of “c”?

Specifically, how does one factor “y – 2” from “- (cxy) + (6x)”?

Thank you for all your help, Mike.

First, an apology: while I aim to answer Q&A questions within 24 hours, I’m late on this one. Sorry!

This is a tricky one, and pretty unlike anything I’ve seen on a real test. For that reason, I wouldn’t worry too much about it. As you note, factoring y-2 out of the first part is easy enough.


From there, you have to get creative. The way I think through it is that I know y-2 must be a factor of -cxy+6x. That means I know (y-2)(\text{SOMETHING})=-cxy+6x.

Figuring out the SOMETHING requires recognizing a few things:

  • It must contain x (both terms on the right contain x).
  • It must contain -3 (the second term gets multiplied by -2 and ends up positive 6x).

So what if we say that SOMETHING equals -3x ?


Does that help?

Hello! Do you think you could explain number 8 in the Polynomials Practice Questions of the PWN the SAT Math Guide 4th Edition? Thank you!

Gladly! I think the best way to go for this one is to factor.



From that, you can see pretty quickly that two answer choices aren’t right. First, you see that g(x) is most definitely a factor of f(x), so A isn’t the answer. You can also see that g(x) has two real zeros (–5 and 2), so C isn’t the answer.

You can also probably see what the answer is. Once everything is factored, it’s clear that (x-5) is not a factor of f(x). That means B is the answer.

As for D, well, once you know the answer is B you might not want to worry about it. But in case it’s keeping you up at night, here’s the deal. If h(x)=f(x)+2g(x), then we can use the factoring we’ve already done to write the following.


See how you can factor (x+5)(x-2) out of both terms?


That’s how you know that (3x+2) is a factor of h(x), which means you can cross off choice D.

Test 1 Section 3 Number 16

One way to go here is to know the difference of two squares: we can factor t^2-4!


That tells you that t must be either 2 or –2.

Since the question told you that t>0, you know the answer must be 2.

You can also get this just by solving more traditionally, and that won’t take much longer.

    \begin{align*}t^2-4&=0\\t^2&=4\\t&=\pm 2\end{align*}

Hi Mike: I get tripped up by factoring Qs like this, especially “NOT” Qs… What’s the best way to solve this? Tks!
Factoring the polynomial x^12 -9 reveals a number of factors for the expression. Which of these is NOT one of the possible factors?

A) x^6 +3
B) x^6 -3
C) x^3 + √3
D) x^3 – √3
E) x – √3

I’m not sure that “NOT” questions are a real pattern of questions in official materials, so I wouldn’t worry too much about this, especially because the presence of 5 answer choices is an immediate giveaway that this is not a question designed for practice on the new SAT.

The way to go here, though, is to factor the difference of two squares multiple times. (The answer choices provide a clue that that’s what you need to do.)

    \begin{align*} &x^{12}-9\\ =&(x^6+3)(x^6-3) \end{align*}

That’s a start–it helps you eliminate choices A and B. From there, look again at the answer choices to guide you further. (Remember, answer choices are 100% part of the question! Use them!)

E looks a bit different than C and D, so I’m leaning that way. How can I factor (x^6+3)(x^6-3) to get to C and D? Well, by using the difference of two squares again, even though 3 isn’t really a square.

    \begin{align*} &(x^6+3)(x^6-3)\\ =&(x^6+3)(x^3+\sqrt{3})(x^3-\sqrt{3}) \end{align*}

Hi, thanks for taking the time to answer my question!

Which of the following sets contain only factors of the number 75?

(A) {1,4,5,20}
(B) {1,3,5,25}
(C) {0,75,100,125}
(D} {3,15,17,25}
(E) {2,3,5,15}

Please explain how you did it as well, thank you 🙂

A factor of x is a number than evenly divides x. So 3 is a factor of 75, because 75/3 = 25, which is an integer. 10 is NOT a factor of 75, because 75/10 = 7.5, which is NOT an integer.

The best way to go for this question is just to list the factors of 75, which you do by dividing it by a bunch of things until you start getting repeats. First divide it by 1 (1 is a factor of everything). Then divide it by 2 (which won’t work, in this case). Then divide by 3, etc. The list of factors you should come up with:

1, 75
3, 25
5, 15

Once you divide by 5, if you keep going you won’t get another factor until you get to 15. Since you already had 15 from when you divided by 5, you know you’re done and have listed all the factors of 75.

Now look at your answer choices. Only choice B has numbers from the list we generated and no other numbers.

If cosA is not equal to 1, then sin^2A/(1-cosA) =

A) 1 + cosA
B) cosA
C) 1-cosA
D) 1
E) cosA -1

Remember the mother of all trig identities:

\sin^2 A + \cos^2 A = 1

The minute you see that \sin^2 A in there you should be thinking of that identity. Rearrange it so that you can do some substituting:

\sin^2 A =1- \cos^2 A

Now here’s what you’re working with:

\dfrac{1- \cos^2 A}{1- \cos A}

Now let’s get nuts and factor the difference of two squares up there in the numerator:

\dfrac{(1+ \cos A)(1- \cos A)}{1- \cos A}

Of course, you can simplify that to just 1+\cos A

It’s exciting times around PWN HQ—lots of things going on. 2014 should be a fun year for SAT prep. That has nothing to do with this contest, of course. I just like to open these contest posts with a little friendly chatter. I bet nobody even reads this stuff. :/ ANYWAY, here’s a challenge question!

For all positive integers n, let ↭n equal the number of unique prime factors of n. For example, ↭12 = 2, because there are 2 unique prime factors of 12: 2 and 3. If a is a positive integer less than 500,000, what is the greatest possible value of ↭a?

Post your answer in the comments—the first correct answer from someone who hasn’t won a contest before wins a Math Guide! Comments are set to require moderation until tomorrow to add a little suspense into the contest.

All the usual contest rules apply. One recent addition to the contest rules I’d like to draw your attention to:

  • No answer changing. When you post a comment, I get an email with that comment, and that’s what I use to judge the contest. Edits applied to your comment later don’t count, even if the edit occurs before someone else wins. Don’t post your comment until you’re sure of your answer.
Good luck!

UPDATE: Wow—I thought this would be tougher for you guys! Congrats to everyone who got it right, especially to Nick, who got his answer in first. Explanation follows below the cut…

The key to getting this question right is recognizing that the smaller the prime factors you use, the more you’ll be able to fit in before your product exceeds 500,000. Therefore, all you need to do to solve this problem is test out the incremental products of the smallest prime numbers!

2 × 3 = 6
2 × 3 × 5 = 30
2 × 3 × 5 × 7 = 210
2 × 3 × 5 × 7 × 11 = 2,310
2 × 3 × 5 × 7 × 11 × 13 = 30,030
2 × 3 × 5 × 7 × 11 × 13 × 17 = 510,510 ← Oops, too big!

What we’ve just shown there is the smallest number that can have 2 unique prime factors is 6, the smallest number that can have 3 unique prime factors is 30, …, the smallest number that can have 7 unique prime factors is 510,510. Therefore, no positive integer less than 500,000 can have more than 6 unique prime factors.

So this isn’t a super important thing as far as how often it appears on the SAT, but it does pop up time and again, so if you’re shooting for perfection (or close to it) you might want to pay attention. Otherwise, you can get by just fine without this little nugget (but you might as well read it, since you’re here anyway).

Do you know what prime factorization is? Basically, the prime factorization of a number is the way you would build that number by multiplying together only prime numbers. To find the prime factorization of a number, divide by 2 if you can. Do that as many times as you can. Once you can’t do that anymore, try dividing by 3 as many times as you can. Then by 5. Then by 7. Then by 11. I think you get the idea.

Let’s try one together, like best friends

What is the prime factorization of 13728?

Whoa. Big number. Lots of people like to make trees when they do this. Let’s do that. Damn I wish you and I were in the same room with a chalkboard right now. This is going to take flippin’ forever.

See how, when I couldn’t divide by 2 anymore, I went to three, and then to 11? I knew I was done when I had two prime numbers, 11 and 13. If I multiplied all those numbers back together, I’d get 13728 again. For serious. Try it:

2 × 2 × 2 × 2 × 2 × 3 × 11 × 13 = 13728

So why do I need to know this?

Because sometimes the SAT asks hard questions (and if you took the October 2011 SAT, you can confirm) about the lowest multiple of two numbers that’s also perfect square. It just so happens that prime factorization is a great way to find a perfect square.

The prime factorization of a perfect square will contain even numbers of each prime number. Look back at the prime factorization of 13728. That’s not a perfect square. There are 5 2s, and one each of 3, 11, and 13. We can use this information to find the lowest multiple of 13728 that’s a perfect square. In order to make a prime number, we’re going to need another 2, another 3, another 11, and another 13. Yikes. That’s gonna be a big number.

2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 11 × 11 × 13 × 13= 11778624

Peep the bold underlines. Those are the factors I’ve added. My new product is huge. It’s also a perfect square. Seriously.

11778624 = 3432

And there are no multiples of 13728 that are less than 11778624 that are perfect squares. Scout’s honor.

What would an SAT question about this look like?

Glad you asked. Try this (no multiple choice — it’s a grid-in):

  1. If p2 is a multiple of both 8 and 35, and p is a positive integer, what is the least possible value of p?

So…yeah. Start by doing a prime factorization of 8 and 35.

8 = 2 × 2 × 2
35 = 5 × 7

Note that you have odd numbers of all three used prime factors. You’re gonna need another 2, another 5, and another 7.

2 × 2 × 2 × 2 × 5 × 5 × 7 × 7 = 19600
p2 = 19600

Confirm that 19600 is a multiple of both 8 and 35 (of course it is):

19600 ÷ 8 = 2450
19600 ÷ 35 = 560

Yes, it worked. So what’s p? Just take the square root of 19600!

19600 = 140
p = 140

Note the tempting false shortcut: just multiply 8 by 35 and square the result. But if you do that, you get 78400 for p2 and 280 for p. That’s not the smallest possible p, as we just showed.

Like I said, you don’t see this often on the SAT, but if you’re shooting for perfection, you’ll want to know this relationship between prime factors and perfect squares.