…Say what now?

This isn’t tested on the SAT all that often, but it has appeared (you’ll find an example in the Blue Book: Test 3 Section 5 Number 8) and I’ve had a bunch of kids tell me lately that they don’t remember ever learning it in school.

When you have two polynomials that equal each other, their corresponding coefficients equal each other.


ax2 + bx + c = mx2 + nx + p


a = m
b = n
c = p

You might find it useful, in fact, when you’re presented with polynomials that equal each other, to stack them on top of each other and put circles (use your imagination because I can’t figure out how to circle things in HTML) around the corresponding coefficients:

ax2 + bx + c
mx2 + nx + p

Example (Grid-in)

(x + 9)(x + k) = x2 + 4kx + p

  1. In the equation above, k and p are constants. If the equation is true for all values of x, what is the value of p?

Alllllright. First, foil the left hand side:

(x + 9)(x + k)
= x2 + 9x + kx + 9k

Might look a little better like this:

x2 + (9 + k)x + 9k

Now stack up the two sides, and see what equals what:

x2 + (9 + k)+ 9k
x2 + 4kx + p

So we know that:

9 + k = 4k
9k = p

From here, this is cake, no?

9 + k = 4k
9 = 3k
3 = k

9(3) = p
27 = p

Math is fun!

Try a few more, won’t you?

You need to be registered and logged in to take this quiz. Log in or Register

Also check out this brutal old weekend challenge testing the same concept.

Comments (21)

The total coefficients of each power of x will be equal. So in the case of 17, once you foil it out, you get:

x^2 – 3xdx + 3d = x^2 – 2dx + m

The important (and confusing) part is:

–3xdx = –2dx

Which simplifies to:

3x + dx = 2dx
3 + d = 2d
3 = d

Once you know that, you can solve for m, which is equal to 3d, or 9.

If d = 3 and m = 9, then dm = 27

Starting from here:
x^2 – 3xdx + 3d = x^2 – 2dx + m

First subtract x^2 from both sides:
– 3xdx + 3d = –2dx + m

Then do a little factoring to have single coefficients for x on both sides:
(–3 – d)x + 3d = (–2d)x + m

Since you know corresponding coefficients are equal to each other, you know 3d = m, and (–3 – d) = (–2d).

Does that help?

I lined up ax^2 – bx + c = rx^2 + sx + t If the equations are true for all values of X, why would B = S not be true? Everything else lines up perfectly and it would work if B = S = 0. Other numbers won’t work because B is negative, so wouldn’t that contradict the statement in the question, that it is true for all values of X? I acknowledge that some numbers won’t work for B = C, but I’m not sure how the rest is true if that is how it is?

Only 0 would work. I understand that, but I still don’t know how the equations can be equal for all values if for any integer besides 0, there will be a -B and a positive S. (A and R are equal as well as C and T) -B and S are not except for the integer 0 . That means the equations are only equal if B and S = 0. Am I even making any sense? I tend to over-think things.

You’re overthinking this one, I’m afraid. If b = –2 and s = 2, then the equations are the same.

The point here is that the WHOLE coefficient, including the sign, are equal if the polynomials are equal. So it’s not that b = s, it’s that (–b) = (+s). Does that help?

I do think so! Thank you Mr. McClenathan. I guess I should just move on next time if I got the answer right. Process of elimination / plugging in did help me. (A) was just the odd one out, except for 0, and it just bothered me.

I took this PSAT and only took the math and reading sections and I am an eighth grader. I don’t know if this test is reliable as I only scored a 610 on the math and 780 on Reading. However, on the SAT in 7th grade, I scored a 1910 (660M, 610V, 640 W) Here is the link (from Mcgraw hill) Please tell me if the math is hard or easy so I can judge my strengths and practice on your awesome blog! www. mhpracticeplus.com/tests/PSAT/0874Ch14d.pdf

I’m not able to go through this whole test and give you a sense of its relative difficulty. When you take non-CB practice tests, you shouldn’t worry about difficulty—the score you get will almost definitely not map directly to what you’d score on a real test. The only value is in identifying concepts that you still struggle with.

Leave a Reply