…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?

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