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Using Synthetic Division to Factor - Problem 1
Factoring polynomials using synthetic division. So in front of me I have a third degree polynomial, I don’t know how to factor this without anything other than synthetic division, I can’t just look at it and say, oh! It’s this times this times this. So what we have to do is use synthetic to see if we can find some factors, okay?
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.