Using Synthetic Division to Evaluate Polynomials - Problem 1

Using synthetic division to evaluate a polynomial function; this particular example we are looking at a pretty straight forward example. We have a quadratic equation and we want to find what happens when we plug in 2.

So this problem is actually straight forward enough that we don’t have to do synthetic division, but I’m going to do it just so you actually see why this works and it’s not this crazy mystery. So first we can just evaluate out as we normally would. This would be 2² minus 3 times 2 plus 4. 4 minus 6 plus 4 is going to give us 2, so that’s easy enough to do. We don’t actually have to do synthetic division to figure this out, but like I said I want do it just so you can see how this all works together.

If we wanted to evaluate this with synthetic division, we first throw the number we have up here on the outside which is 2 and then write our coefficients down in descending order 1, -3 and 4. Using synthetic division drop down the first term, multiply and then add 3 plus 2 -1, -2 leaving us with a 2. So the trick for this is interpreting our results.

So what we actually did is we put 2 on the outside. Now when we use synthetic division think back when you were actually dividing, so we said what we’re doing is x minus 3x plus 4 divided by something. The number we put on the outside was actually the number that made the denominator 0. Again when doing synthetic division we can’t have any coefficients on that term, so really what this is telling us is this had to be x minus 2. By doing synthetic division with 2 this is actually what we evaluated.

So looking at our answer, if we’d evaluated this out this would be our remainder which is the answer we’re looking for by the way in case you missed that, and this is going to be x minus 1. So I want to take it over to the other side of the board and look at actually what we have altogether. So what we did is by doing synthetic division we said that x² minus 3x plus 4 over x minus 2 is equal to x minus 1 plus our remainder over our divisor.

So really that’s why we did synthetic division and we found this out. If we multiplied both sides by x minus 2, then what we’re left with x minus 2 over here it goes away. X² minus 3x plus 4. The x minus 2 gets multiplied by the x minus 1. I don’t really care to multiply that out, and the x minus 2 cancel here just leaving us with a plus 2.

What we were trying to find was p(2). This is p right? So what we have when we plug in 2 is what comes out of here. These are equivalent statements though, so we’re going to get the same thing when we plug 2 into this side as we do as we plug 2 into this side, but the most important piece is this right here. What happens when we plug 2? 2 minus 2 this disappears, so all we’re actually left with is this little piece of remainder, so p(2) is what happens when we plug in 2 over here and what we’re left with over here is just the remainder or 2.

So in essence what I just wanted to show you over here is why synthetic division worked to evaluate the function. It’s not a mystery to evaluate yeah evaluate functions. It’s not a mystery it’s just sort of using the intricacies of synthetic division and how everything is related in order to get something to cancel out and give us what we want.