We just started hiking up Polynomial Mountain, and we've already found it. The Remainder Theorem.
No worries—we know its name sounds scary. We hear "remainder" and think "division" and shudder, but this is actually another little trick showing us how to evaluate polynomials for a specific value. And we love little tricks—right?
More good news: if atmospheric conditions are right, this is also a way to avoid long division, and a whole lot of headaches altogether.
Hang on. Our polynomial friends are clearing their throats. They want to say something to us.
If we divide a polynomial, P(x), by x – a, then the remainder equals P(a).
That must be it: the secret of The Remainder Theorem. They've entrusted it to us. Let's give it a whirl.
Sample Problem
Find the remainder when 3x2 – 4x + 5 is divided by x – 1.
According to our brainy polynomial friends, we need to plug x = 1 into our expression. Remember, we're dividing by x – a, so we need 1, not -1. Using the wrong sign would teleport our answer straight to Wrongville. Not good.
P(1) = 3(1)2 – 4(1) + 5
= 3 – 4 + 5
= 4
The remainder is 4. Now we know that x – 1 does not divide evenly into 3x2 – 4x + 5.
The polynomials are chittering away and getting excited. They want to show us another example. We'd hate to disappoint them.
Sample Problem
Find the remainder when 2x2 + x – 7 is divided by x + 2.
We start by plugging -2 into our polynomial.
P(-2) = 2(-2)2 + (-2) – 7
= 2(4) – 2 – 7
= 8 – 9 = -1
Okay, the remainder is -1. This isn't too bad.
Uh oh. They're bringing us another problem to solve. They're really into this. We can do one more, but that's it.
Sample Problem
Find the remainder when 2x3 – 2x2 – 2x + 2 is divided by x – 3.
We like this polynomial, so we're gonna go ahead and stick a 3 on it.
P(3) = 2(3)3 – 2(3)2 – 2(3) + 2
= 2(27) – 2(9) – 6 + 2
= 54 – 18 – 4
= 32
Okay, the remainder is 32, and we're done now. That's the end of—wait, they're bringing another problem. Make a break for it.