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Using Synthetic Division to Evaluate Polynomials - Problem 2
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Using synthetic division to evaluate a polynomial; so for this particular example we are looking at a third degree polynomial, a cubic polynomial, and we’re asked to find p of -2.

So we could just plug it for all the xs and just see what we’ve come up with, but what I want to do is explain how we can use synthetic division to do the same exact thing. So what we do is we make a giant bracket, the number that goes on the outside is the same exact number as we are trying to plug in in the first place. So we put -2 on the outside and then in descending order, we write the coefficients of the polynomial we’re looking at so in this case we have 1, 6, 3 and there is no constant term in this polynomial so we need to throw in a 0 as a place keeper, every single degree needs to be represented.

From here we drop down our first number, 1 multiply -2 times 1 and our answer goes below the next number and add so 6 plus -2, 4 and then continue. -2 times 4 is negative 8, 3 plus -8, -5, -2 times -5 is 10 and our last number is 10.

When we do this, the only thing we’re actually concerned with is the remainder. We go through this whole process just to get this last number and that is actually our answer. Pretty easy no taking big powers just simple addition and multiplication in order to evaluate the polynomial.

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