 ###### 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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# Using Synthetic Division to Solve an Equation - Problem 1

Carl Horowitz ###### 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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Using synthetic division to solve a larger polynomial. So for this particular example we’re going to be looking at a fourth degree polynomial, a really big polynomial that we want to factor. However, we have the good luck of having a list of potential possible zeros. So knowing what we already know we could go through the rational root theorem and find out all our potential zeros, factors of the last term over factors of the last and sort of go through a process of elimination figuring out which factor actually works.

However, in this case we’re given some options. We know that either 2 or -2 are the root, we also know that 3 or -3 is a root. So instead of having to go through everything, we can just focus on those four numbers, okay? So it doesn’t really matter where we start, we just want to start somewhere and pick one of these numbers and see if it works.

I’m going to try -2, why not? So I want to try a synthetic division with -2 and see what comes out. The main thing we want to focus on is if our remainder is 0. So we’re doing synthetic division, -5, -11, 20 and 12. We’ll go through the process, see what comes up. 2, -4, -9, 18, 7, -14, -6, 12, okay. And remember that we are actually adding so this is going to end up being 24. We have a remainder which tells me that -2 is not a 0, is not a root of this particular example. So what that tells me is I can cross that off, it’s not going to work.

So we now know that 2 has to be a 0. So we can go back and do synthetic division again. Instead of rewriting everything, I’m just going to erase the previous work that we just did and take away my negative. The same exact problem, this time trying it with 2.

Going through the same process again, just going through the synthetic division. -2, -13, -26, -6, -12, 0, okay. So 2 is in fact a root, and the important thing is to think back and, if this is a root, this tells me that x minus 2 is a factor. Remember that you always plug in whatever makes a 0, so in this case we plug in 2, this will be 0 that’s going to go along.

So we used our 2 or -2 fact, now we have to go over to the 3 or -3. A common mistake is to restart the process, go back to our initial equation. But what we’ve already found is that this is going to be the quotient when we divide by this. So if 3 is a factor of this it also is going to have to be a factor of this, our resultant. So instead of going back to our original equation, we can use our result piece right here, okay? Again, choosing between 3 or -3 it doesn’t really matter. We’re just looking for a remainder of 0. I’ll try 3, see what happens.

And I just like to draw my bracket right on top of the other one saves a little bit of writing. Going through the synthetic division process, drop down the 2 and 6, 5, 15, 2, 6 and 0. Cool, so 3 is a factor of 0 as well. Again, going through the same logic as this, this is telling us that x minus 3 is a factor.

So what we found is, we have x minus 2, we have x minus 3 and times our quotient over here. So let’s rewrite this over here, let’s get a little bit more room. What we have found is that this exact equation is the same thing as x minus 2, x minus 3 times 2x² plus 5x plus 2. So we’ve done synthetic division twice to take a forth degree polynomial back down to a second degree polynomial which we know how to factor. This we know we both have to be pluses, and our 2x has to be paired with our 2 to make 4 to have it work. So this is going to be equal to x minus 3, sorry, x minus 2 times x minus 3 times our factored quotient.

So we have factored this completely, okay? If we wanted to find these zeros we now just look and see which term, what value of x would give us 0 out of each term and we have solved this equation out.

So using synthetic division to solve a polynomial; in this particular incidence we’re given some restrictions so we know what values to try but we could also use the rational root theorem and just plugging and chugging figure out what gives us a remainder of 0 to get to the same results.