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Using Synthetic Division to Factor - Problem 2
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!
Factoring a polynomial using synthetic division. So for this particular problem behind me, we are trying to factor a third degree polynomial. We can’t do that by hand because we don’t know how, we don’t have the skills to do that, but we can if we have a piece of information, in this case we know one factor.
So what we really want to do is figure out what this polynomial divided by the divisor is and then we can factor that result if everything goes well. So in order to divide this out we just use synthetic division. Take whatever it gives our divisor a zero, so that’s going to be 3 in this case, and then write out all our coefficients on our dividend, in this case, 2, -5, -4, and 3. There’s no powers that are missing so we don’t need any zeros to hold spaces.
And then we drop down our first term, multiply and add. Keep on going and we end up with a remainder of 0 which is a good thing because we were already told that it’s a factor. If we had something else, we would have known something was wrong.
So what we actually just found out then is our dividend, our cubic, we have y² minus 4y plus 3, divided by y minus 3, is equal to this piece right here 2x² plus 1 minus 1. Remember that this is our remainder, our constant term, our x term, I forgot my x, and our squared term.
So if we’re trying to factor this we can actually multiply this y minus 3 around. So we end up with this times y minus 3 and then we end up with our third degree polynomial equal to this, our quotient times our divisor.
From here we just want to factor, did I change my Xs and my Ys? I did, I have a tendency to do that from time to time. These should all be Ys over here, I apologize for that, but the variable is the same thing, okay.
So from here what we want to do is, we have it down to a quadratic, just factor as we would anything else. And we still have this y minus 3 on the end. We know that we have a plus and a minus because we have a minus and we know that our plus is going to have to be bigger, which tells me that this is going to be a 1 and that’s going to have to be a 2y. 1 in both places, sorry that would be a y over here as well. Checking to make sure our term is marked up, I’m going to have 2y² plus 2y minus y plus y, minus 1.
So what we did is using synthetic division, we turned our third degree into a second degree which we then know how to factor, factored it up living us with our factored answer.
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