##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Using the Conjugate Zeros Theorem - Problem 1

###### 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.

I want to find the zeros of a polynomial function using the two theorems, the one that says that a polynomial has as many zeros as it's degree, and the other the conjugate zeros theorem that says that imaginary zeros come in conjugate pairs.

I have this polynomial f of x equals f to the fourth minus 6x³ plus 14x² minus 24x plus 40 and I know one thing about it that f of 2i is 0, that means that the number 2i is a zero of this function.

I want to find all the other zeros using this fact. Well that first theorem, the theorem about the number of zeros tells me that I'm going to expect four zeros counting multiplicity for a fourth degree polynomial function and the second theorem, the Conjugate Zeros Theorem tells me that if 2i is a zero, then minus 2i it's conjugate, it's also a zero, so 2i and -2i are zeros and that means by the Factor Theorem that x minus 2i and x plus 2i are factors of this polynomial so let me multiply this through.

This is actually a difference of squares, the nice thing about conjugate zeros is they always give me a difference of squares, so this is x² minus the quantity 2i squared.

Now 2i squared is -4, so x² minus -4 is x² minus 4. Now what I need to do, I now know that these two factors and plus this is a factor of my polynomial f of x, I have to find out what the other factors I have to divide and this is going to be pretty serious division problem. I can't use synthetic division because this is not the right kind of divisor, synthetic division only works for divisors like this x minus 2i.So I'm going regular old polynomial division for these two polynomials.

So I've got x² plus 4 and x to the fourth minus 6x³ plus 14x² minus 24 x plus 40 all right now what do I need to multiply by x² to get x to the fourth? X² and I multiply through x to the fourth and 4x², I put the x² column, I change signs and add. This gets 0, -6x³ and plus 10x² and let me pull down the -24x.

Now what will I multiply by x² to get -6x³? -6x, I multiply through -6x times x², -6x³ by design and -6x times 4 is -24x, I change the signs and add and I get 0, 10x², 0 and I pull down the 40. Perfect right? This is exactly 10 times this and so I need to multiply this by 10 to get this and I multiply through 10x², 40 change the signs and add and I get 0, remainder 0 tells me that this was indeed a factor of this polynomial.

Now what this means is that my function can actually be written as x² plus 4 times, this other factor x² minus 6x plus 10. Now this other factor hides the other zeros, I found 2i, I know that 2i and minus 2i's are zeros based on this factor, but this guys hides the other two. Let me use the quadratic formula to find the other two. X equals -b, b is -6, so 6 plus or minus the square root of b², 36 minus 4ac, so minus 40 all over 2a which is 2. Now this is the square root of -4 which is 2i, 6 plus or minus 2i over 2, that's going to give me 3 plus or minus i and that's it. Those are the two remaining zeros.

My zeros are my number from the beginning 2i, its conjugate -2i, 3 plus i and 3 minus i. This polynomial has no real zeros if you would have graphed, it wouldn't cross the x axis at all, but it's got four zeros just like the theorem predicts same number as it's degree and the imaginary zeros come in conjugate pairs.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete