# Finding Zeros of a Polynomial Function - Problem 1 5,397 views

I’m finding the zeros of polynomial functions, let’s take a look at this fourth degree polynomial function a quartic function and I want to find all its zeros. And remember the first step is to find the potential zeros using the rational roots theorem.

The way that works is we look at the integer coefficients of 5 plus or minus 1 plus or minus 5. For example -1 times -5 is 5 and the integer coefficients of 2, plus or minus 1, plus or minus 2. And we’ve form all the possible fractions of these over these right, the coefficients of the constant over the factors of the leading coefficients.

So you’ve got plus or minus 1 plus or minus 2 over plus or minus 1 plus or minus 5. That’s a lot of possibilities. I always like to start with plus and minus 1. 1 over 1 or -1 over 1. Start with those two.

Now let’s, start with 1. We'll use synthetic division to test this. You’ve got 5, 4, -11, -8, 2 I’m just copying these coefficients down here. And then drop the 5 down and multiply by 1. You get 5, add and multiply add and multiply, add and multiply and I get -10.

This final answer -8, this is the remainder that I get when I plug 1 to this function. So this is actually f of 1, -8. So it's not 0, therefore 1 is not a 0 of this function. Let’s try a -1.

The same coefficients 5, 4 -11, -8, and 2. Bring 5 down multiply by -1, add, multiply by -1, add, multiply by -1 and add I get 2 times -1, -2 bingo. I found a zero, -1 is a zero. That means that this function can also be written as x plus 1 times and this is the reduced polynomial, 5x³, it's got 4 terms this is a constant x term, x² term, x³, 5x³ minus x² minus 10x plus 2.

So I will write that down, minus x² minus 10 x plus 2. Now I need to keep checking for more zeros I have a cubic function left over. So let’s take a look we’re still going to come for this family of numbers here let’s try 1/5.

So I’m going to use this polynomial now not the first one it will be a little bit easier. So 1/5 is the number I’m going to try coefficients of 5 -1, -10, and 2. I drop the 5 down and multiply, add and multiply, add and multiply by 1/5. 1/5 times -10 is -2 bingo I found another one.

1/5 is another zero and here is my reduced polynomial and now I have that f of x is x plus 1 times x minus a fifth. And then the remaining polynomial is 5x² minus 10. Now I get a factor of 5 here and I get a 1/5 here. I can get rid of the fraction just by moving the factor of 5 into this factor x plus 1 5 x minus 1. I knew these things up a little bit and I’m left with x² minus 2. Super easy where are the zeros of x² minus 2?

You got x² equal to 2 so x is plus or minus root 2 those are the remaining zeros, and so the zeros are -1, 1/5 root 2 and negative root 2.