Brightstorm is like having a personal tutor for every subject

See what all the buzz is about

Check it out

Finding Zeros of a Polynomial Function - Problem 2 3,800 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s find the zeros of another polynomial, this time f of x is equals to x to the fourth minus 9x³ plus 13 x² minus x minus 5. Now the first step is to find the potential zeros using the rational roots theorem.

So I need to look at the integer factors of the leading coefficient and the integer factors of the constant, like 1 times -5 could give me -5. And then I form all the possible fractions of these numbers divided by these numbers. So plus or minus 1 plus or minus 5 over plus or minus 1 plus or minus 2.

I’m going to start testing these numbers using synthetic division. And I always like to start with 1 and -1 just because arithmetic’s is easy and you make it lucky. So let’s try 1 first.

So 2, -9, 13 -1 -5. Bring the 2 down multiply -7 you add these and multiply, add and multiply, add and multiply, right off the back we get lucky. That means that f of 1 is 0, so 1 is a 0. And it also tells us that when we divide x minus 1 the factor that corresponds to this 0, when we divide it out we get this reduced polynomial 2x³ minus 7x² plus 6x plus 5 so let me right that up over here. 2x³ minus 7x² plus 6x plus 5.

Now the rest of the zeros are going to be hiding in that cubic but they are still among this family of numbers so I haven’t tried negative -1 yet why don't I try that. -1, I will use this polynomial instead it will be a little easier. Always easier to use the reduced polynomial. Drop the 2, multiply by -1, add, multiply, add and multiply and unfortunately negative term we didn’t get lucky this time.

This means that if I plug -1 into this polynomial I get -10 it’s not a zero. So let’s start looking at the fractions. How about 1/2? So try 1/2 as a zero this polynomial. 2, -7, 6, 5, drop the 2 and multiply to get 1. Add them -6 multiply -3, add, get 3 and multiply 3/2, 5 and 3/2 doesn’t really matter what it is, it's not 0. So 1/2 is not a zero of this polynomial. Let’s try -1/2.

-1/2 we have 2, -7, 6 and 5 we drop 2, multiply add multiply add and multiply 10 times -1/2 is -5, perfect, we found another one. -1/2 is a zero. And that means we can right x minus 1, x plus 1/2 and the reduced polynomial 2x² minus 8x plus 10. And I always like to neaten things up.

I have an extra factor of 2 here that I can pull out put into this guy that gives me x minus 1, 2x plus 1, x² minus 4x plus 5. I always find it easy to factor out find zeros if I get rid of the common factors. Well we have a quadratic, so all we have to do is use the quadratic formula to find the zeros of this guy.

So x equals -b, -b is negative -4, positive 4, plus or minus b², -4 square is 16. Minus 4ac 1 times 5 times 4, 20, so minus 20 all over 2 a which is 2. This is -4 and the square root of -4 is 2i, it's imaginary. So 4 plus or minus 2i over 2 and that reduces to 2 plus or minus i.

We have two imaginary solutions. So our zero’s are 1, -1/2, 2 plus i and 2 minus i.