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# Using the Conjugate Zeros Theorem - Problem 3

###### 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 take a look at an extra hard problem of finding the zeros of a polynomial.

Here I've got a fifth degree polynomial function and I'm told that f of -½ plus root 3 over 2i is zero and f of 1 minus ½i is zero. We're going to use the Conjugate Zeros Theorem. If you know that -½ plus root 3 over 2i is a zero, then you also know that -½ minus root 3 over 2i, it's conjugate. It's a zero and that means that x minus this is going to be a factor. That means x plus ½ minus root 3 over 2i and x minus this is a factor of our polynomial. X plus ½ plus root 3 over 2 that's a 3 over 2i.

Now remember our trick for multiplying these special factors when you have conjugate imaginary zeros. The x plus ½, you have in both factors and the root 2 over 3i you have in both factors, here you have a minus, here you have a plus so what we have here is actually a difference of squares and if you use that it will save you a couple of steps of multiplication. We have x plus ½ quantity squared minus root 3 over 2i quantity squared.

Now expanding this, this is just a binomial we get x² plus ½ x times 2 plus x plus ½ squared is a quarter and then I have root 2 over 3i squared, root 3 over 2 squared is ¾, root 3 over 2i squared is minus ¾ so I have minus -¾ plus ¾. So this is just 1. X² plus x plus 1. Now I've got to divide that quadratic put of my fifth degree polynomial and find what the remaining factor is, so I give myself a lot of space to do the division x² plus x plus 1 divided by and my polynomial is 8x to the fifth minus 20x to the fourth, plus 14x³ minus 9x² plus 19x minus 15. It's going to be a good one.

All right, what do I have to multiply by x² to get 8x to the fifth? 8x³, so I multiply through and I get 8x to the fifth plus 8x to the fourth plus 8x³, I change the signs and I add and I get 0, I get -28x to the fourth, I get +6x³ and I'll pull down the minus 9x².

What do I have to multiply by x² to get -28x to the fourth? -28x², I multiply through and each of these has a coefficient of 1, so I know what I'm going to get -28x to the fourth, minus 28x³ and minus 28x², change the signs and add. I get 0, I get 34x³, I get 19x², what do I need to multiply by x² to get 34x³? 34x and I multiply through 34x³ plus 34x² plus 34x and I need to bring down this 19x. I change the signs and add and I get 0, I get what? -15x² and -15x and thankfully I've got a -15 and I can pull down, that's perfect, so what do I multiply by x² plus x plus 1 to get this? -15 and just to conserve space I'm not going to check that.

So this is my quotient, that means my polynomial can be factored into x² plus x plus 1 and 8x³ minus 28x² plus 34x minus 15, but we're not done. We haven't used the fact that 1 minus 2i is a factor and if 1 minus 2i is a factor, since this has all real coefficients, 1 plus ½i is going to be a factor, a zero let's use that fact now.

So 1 minus ½i and 1 plus ½i are zeros, that means x minus 1 plus ½i and x minus 1 minus ½i are factors so I multiply this through remember the trick x minus 1 quantity squared minus ½i quantity squared and so I get x² minus 2x plus 1 and I have minus a quarter, so minus, minus a quarter is plus a quarter. This is x² minus 2x plus 5/4.

Now let's notice that our polynomial here, we have an 8x³ minus 28x² plus 34x minus 15 and this is one of its factors, so if I divide this factor out of the cubic, I'm going to get the one remaining linear factor, so I have to do that division now. X² minus 2x plus 5/4 divided by 8x³ minus 28x² plus 34x minus 15.

Let's go in, so what do we have to multiply by x² to get 8x³? 8x, multiply through and I get 8x³, 8x times -2x is -16x², this is lucky right? We have 5/4 times 8, 8 and 4 cancel giving me 2, I have 10x so plus 10x, change the signs and add and I get 0, I get -15x², I get 24x and I get -15.

Okay this is the moment of truth, what do I have to multiply by x² to get -12x²? -12, multiply through and I just need to check -12x², -12 times -2 plus 24x, -12 times 5/4, the 4 and the 12 cancel giving me 3, -3 rather times 5 perfect.

The final factor is8x minus 12 and that means x² plus x plus 1, remember that this factored into the quadratic x² minus 2x plus 5/4 and the remaining factor was 8x minus 12.

So let's list the zeros. First we had minus ½ plus root 3 over 2i and minus I'll write that this way, plus or minus root 3 over 2i and then back over here we had 1 minus ½i, one plus ½i, so 1 plus or minus ½i and one more, what's the zero of this? 8x minus 12 equals 0, 8x equals 12, x equals 12 over 8, 3/2, that's the last one and that's it.

This represents five zeros right? We had a fifth degree polynomial function, we have a pair of conjugate imaginary zeros here, another pair here and one real zero.

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###### 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!”

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