We're going to be finding the roots of a lot of polynomials before we're done here. We'll have so much experience finding roots, we could start a new life as a potato farmer. How about we review two of our greatest tools for finding the factors of polynomials—they lead us straight to the zeros. Our tools are a spade and trowel. That, or the Remainder Theorem and synthetic division.
Sample Problem
Find the remainder when 2x7 – 5x6 + x5 – 7x4 + x3 + x2 – 1 is divided by x – 3.
The Remainder Theorem says that, when we divide a polynomial by x – a, the remainder from that division will equal f(a). We just plug and chug that 3 in and go. If we're lucky, the result will be 0, and so 3 is a root of the polynomial. If we're unlucky, the earth will open up and swallow us, or we'll find a 2-leaf clover, or a flower pot will drop on our heads, or…
Let's just use the Remainder Theorem before we freak ourselves out over all the bad things that could happen to us.
f(3) = 2(3)7– 5(3)6 + (3)5 – 7(3)4 + (3)3 + (3)2 – 1
= 4374 – 3645 + 243 – 567 + 27 + 9 – 1
= 440
Thank goodness for calculators, or we'd have spent way too long multiplying those exponents out. At least we know 3 isn't a root of the polynomial.
Sample Problem
Divide y = x3 + 3x2 − 4 by x + 2.
We'll be honest: long division works. It's friendly to its neighbors and pays its taxes on time. But if we can get out of doing it, we will. That's where synthetic division comes in.
We set up our box o' division, with our a, -2, on the outside. Inside the box are the coefficients for each term of the polynomial (even the ones that have a 0, like x).
We take the first term down, then multiply it by -2, and then put the result back in the box. Don't worry, there are toys in the box. It likes it there.
Now we add the 3 and -2 together, getting (surprise) 1. Now we start all over again, multiplying that by -2, putting it in the next column, and repeating things over and over until we reach the end of the polynomial.
Success, it divides clean through. The quotient from division is found from the bottom row: starting at the right, we have our remainder (0 this time), then our constant term (of degree 0), and then the degree increases going to the left. So our result is x2 + x – 2. We can now rewrite the original polynomial as (x – 2)(x2 + x – 2).
Note that both the Remainder Theorem and synthetic division only work when dividing by a degree 1 polynomial, with a coefficient of 1. Bummer. Guess there's still a reason to whip out long division from time to time.