Finding Zeros of a Polynomial Function - Problem 3
We're finding the zero as a polynomial functions, let’s try a harder example a fifth degree polynomial x to the fifth plus x to the fourth minus 5x cubed minus 15x² minus 16x minus 6. Now remember we only start by using the rational root’s theorem to find potential zeros.
And that means I have to look at top integer factors of the leading coefficient, plus or minus 1, and integer factors of the constant plus or minus 1, plus or minus 2, plus or minus 3 plus or minus 6. And they form fractions of these over these so I get plus or minus 1, plus or minus 2, plus or minus 3 and plus or minus 6 all over plus or minus 1. So basically these numbers in the numerator these are the numbers I had to check as potential zeros.
Now usually I start with 1 and -1 because they are so easy to do arithmetically. And I want to show you a trick. You can always tell if 1 is a 0 of a polynomial if the coefficients add to 0. Why would that be? Well if you plug 1 is not a polynomial you get 1 over fifth, 1 to the fourth, -5 times 1 all these x terms are going to become 1. So you have 1 ,1 -5, -15, -16, -6.
So you can see that I’m going to get 1 plus 1, 2 and then minus just a bunch of stuff. It’s not going to end up being 0 so 1 is not going to work as a 0. Let’s try -1.
So I use synthetic division to check the rest of these numbers. I write my coefficients 1, 1, -5 , -15, -16 and -6. I bring the 1 down and I multiply add, and multiply, add and multiply, I get -10 adding I multiply, add I get -6, and multiply I get 6. -6 and 6 is 0. Now this is going to be f of -1, I found a 0. -1 is a 0 and that means that x plus 1 is a factor. Now what’s the other factor?
Well it comes from the reduced polynomial here. It's x to the, well what is this constant x, x², x cubed, x to the fourth minus 5x² minus 10x minus 6. Now I have to search in this reduced polynomial for the remaining 0’s,so let’s take a look. And by the way we haven’t changed what is happening in an example yet but it is conceivable that -1 could be a 0 more than once just as it's conceivable that x plus 1² could be a factor of this polynomial. So let’s check that.
1, 0, -5, -10, -6, I bring the 1 down and multiply add, and multiply, add and multiply, I get -6 times -1 is 6 and it does work, it works again. So -1 is 0 twice and that means that x plus 1 is a factor twice. So we have x plus 1² and the remaining factor is this reduced polynomial; x³ minus x² minus 4x minus 6. And it’s here that we'll look for the remaining 0’s.
Now we are looking at these numbers, we've tried plus and minus 1, let’s try 2. Here are my coefficients 1, -1 -4 and -6. We bring the 1 down and multiply, add and multiply, add and multiply and it doesn’t work, -10. So that means when I plug 2 into this polynomial I get -10. It's not a zero that’s okay, let’s keep looking.
Let’s try 3. Same polynomial 1, -1, -4, -6, bring the 1 down and multiply, add and multiply, add and multiply and I get -6 plus 6, 0. So 3 is a 0 of this polynomial and that means x minus 3 is a factor and what’s left is x² plus 2x plus 2.
Now it’s always refreshing when you finally get down to the quadratic because it’s easy to find the zeros of the quadratic, use the quadratic formula. X equals -b that’s -2, plus or minus the square root of b² that’s 4, minus 4ac, a and c are 1 and 2, so 4 times 1 times 2, 8, all over 2a, which is just 2 in this case. We have -2 plus or minus, 4 minus 8 is -4 and the square root of that is 2i it's imaginary, all over 2.
So I get -1 plus or minus i. These are my remaining zero’s. Epic problem, we’ve got -1 is a zero twice, you don't have to write it twice, 3, -1 plus i and -1 minus i. Those are the zeros of my quintic polynomial.