# Using the Conjugate Zeros Theorem - Problem 3 2,144 views

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.