Carl Horowitz

**University of Michigan**

Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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So the first thing I’m looking at is, is x plus 4 a factor? Remember how to use synthetic division, we look at our factor and we see what makes that zero, so that in this case is going to be -4. We draw a giant bracket and then fill in the coefficients of our dividend and making sure we have every single coefficient accounted for. We’re missing an x², so I need to have a place holder of 0, -13, -12.

Synthetic division, we drop down our first term, multiply and then add. And then just rinse and repeat throughout, okay? So -4 times -4, this becomes 16, add 3, -12, now make sure we’re adding here. So this is going to be -12 plus -12 which is -24. So what that tells us is we did synthetic division and we have a remainder, okay? Whenever you divide and you have a remainder it tells you something is not a factor. So is x plus 4 a factor of this? No. Let’s try another example.

Is x plus 1 a factor? Going through the exact same process, we do synthetic division. X plus 1, this tells us we’re going to use -1 on the outside and then doing the same exact process, we have the same polynomial so we’re going to have the same top row.

1, 0, -13, -12 and once again going through the synthetic division process, dropping down the 1, multiplying -1 times 1, -1, and then adding, -1 , 1, -12, 12 and 0. So this tells us that x plus 1 is a factor because we have a remainder of 0. More specifically what that tells us is that this polynomial over here is actually equal to, sorry, let’s take a step backwards.

What we did is we actually divided x³ minus 13x minus 12 divided by x plus 1 and that’s going to equal this polynomial here. This is our remainder, this is our constant term, our x and our x².

X² minus x minus 12. Okay, and what we can actually do is in order to factor this polynomial all together, we could cross multiply. So what we found there then is x³ minus 13x minus 12 is equal to x plus 1 times x² minus x minus 12. And then we now have a quadratic which we can factor by hand. This becomes x minus 4, x plus 3 and this x plus 1 is still there.

So what we did is, using synthetic division we figured out what a factor was and then once we got it down to a quadratic we were able to factor out the rest of the way.