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Finding Zeros of a Polynomial Function - Problem 2
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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.

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