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# Using Synthetic Division to Solve an Equation - Concept

###### 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!

To solve a polynomial equation using synthetic division, we first use the rational roots theorem to determine the potential zeroes for factoring. After factoring, we can solve **synthetic division polynomials** by setting each of our factors equal to the other side of the equation and solving.

Using synthetic division and the rational roots theorem to factor a larger degree polynomial so right here we have a third degree polynomial that I want to factor. Okay so the first thing we have to do is do the rational roots theorem to figure out what our potential zeros are. Okay, factors of last term over factors of the first term positive or negative so we know that we have plus or minus factors of 6 are 1, 2, 3 and 6 and factors of our first term, first term is just 1 so this is just going to be over 1 so we know that our only potential zeros are positive negative 1, 2, 3 or 6 so what we can actually do is do synthetic division with all 8 of these numbers until we get or start doing synthetic division with these 8 numbers until we get a remainder of 0 and we know we have a factor.

Typically that's going to be pretty involved in a lot of work so what can happen is a couple of things. First what your teacher may do is say factor this knowing that only a couple of these maybe real potential zeros okay so they're going to eliminate some of these options for say, instead of having to check all 8 you're down to checking 2 or 3.

The other thing you could do is graph them okay so put in your graphing calculator and a graph for this particular example comes up something like this. Okay so here is our possible rational zeros but then we also could look at our graph and say okay what are our realistic zeros okay looking at this now we could say okay negative 1 doesn't make any sense because there's no zero around here. Positive 6 doesn't make any sense because there's not a zero here so we can use process of elimination to figure out what are more realistic zeros could be making our work significantly less than having to plug everything in.

Okay so say we look at this and you know this looks like 2. We can do synthetic division with 2 then to simplify this down. Let's go try that out, so we do synthetic division with our number 2 and then making sure we write all the coefficients up 1, 0, -6 and oops 1, 0, -7 and then positive 6 okay and do synthetic division drop our 1 down multiply and add 2, 4, -3, -6 and 0 okay so 2 actually did work that's really cool so what we can do then is interpret these results.

What we just found is our quotient to be x squared plus 2x minus 3 and what we did was we divided this by this term so what we have to then include is this is going to be equal to our original polynomial. Part of it is in in the main challenge that we have is we divided by 2 remember how we did synthetic division is we plugged in the number that makes our factor zero so by plugging in 2 this is actually doing division by x-2 okay remember the opposite dividing by if we're dividing by x-2 we put the 2 here. Okay so this statement here is exactly the same thing as up here. We can continue our factoring this turns into x+3, x-1 we still have our x-2 and then finishing it out, this is 2, -3 and 1, remember whenever we're multiplying each term has to 0, so these are our 3 real answers.

Looking at our graph, it makes sense that I looked at the wrong one this is actually 1 but this will be 1 this will be 2 this would be -3 so using our rational roots theorem we found our potential zeros you could either graph plus the combination or your teacher can help you in sort of figuring out where to go from now using synthentic division to solve all that up.

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###### Carl Horowitz

B.S. in Mathematics University of Michigan

He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his step-by-step explanations are easy to follow.

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