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Using the Conjugate Zeros Theorem - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I'm using the Conjugate Zeros Theorem to find zeros of the polynomial function. Let's take a look at this polynomial, this fourth degree, f of x equals 2x to the fourth minus 7x³ plus 7x² plus 73x minus 195 and I'm given that f of 2 plus 3i equals 0, so I want to find the other zeros.
Well if 2 plus 3i is a 0, then 3 minus 3i is a 0 as well because 2 minus 3i is the conjugate of 2 plus 3i and that's the conjugate zeros Theorem, imaginary zeros come in conjugate pairs and that means that we have two factors x minus 2 plus 3i and x minus 2 minus 3i and so I can figure our what quadratic this gives me and divide it out in this function and find the remaining zeros and it's other quadratic factor.
All right well multiplying this looks like it could be pretty hard, but me analyze this a little bit. First of all when you distribute the minus sign here, you get x minus 2 minus 3i and here you get x minus 2 plus 3i. Now let me show you something this is a little trick and it will make things go a little faster. This x minus 2 is in both factors and so is the 3i only here it's a minus and here it's a plus.
This is a difference of squares and when you multiply you can write equals x minus 2 quantity squared minus 3i quantity squared and that makes it a lot easier to multiply it out, so this is going to be x² minus 4x plus 4 and then minus 3x² is -9, so minus -9 plus 9 and we get x² minus 4x plus 13, so that's going to be one of the factors of this polynomial.
To find the other factor, I need to do a polynomial division so let's get started with that. X² minus 4x plus 13 that's the divisor and then my f of x goes inside the division symbol sorry, 2x to the fourth minus 7x³ plus 7x² plus 73x minus 195.
Okay let's buckle down this is going to be a long one. What do we need to multiply by x² to get 2x to the fourth? 2x² multiply through and I get 2x to the fourth. I get -8x³ and I get 26x², change the signs and add. I get 0, I get x³, I get 7x² minus 26, that's -19x² and I want to pull down the 73x.
What do I have to multiply by x² to get x³? Just x, so I multiply through and I get x³ minus 4x² plus 13x, I change the signs and add. I get 0 here, -15x² and 73x minus 13x, 60x, let me pull down the minus 195, that looks a little scary, but actually it will turn out pretty nice. What do I have to multiply by x² to get -15x²? -15 and let's hope this works.
I get -15x², -15 times -4 60x, so far so good and -15 times -13, 15 times 15 is 225, this will be 215 or 30 less, 195 it works, minus 195, change the signs and add and I get 0. My remainder 0 tells me that these are both factors by function and I can write f of x equals my divisor x² minus 4x plus 13 and my quotient 2x² plus x minus 15.
Now I already found two zeros from this quadratic, the two remaining zeros are hiding in this guy and I use the quadratic formula to find them. So x equals -b in this case b is 1 plus or minus the square root of b², 1 minus 4ac, now ac is -30 minus 4 times -30, 120 all over 2a over 4.
So this is 121, the square root of that is 11, so this is -1, plus or -11 over 4. Now -1 minus 11 is -12 over 4 is -3, or -1 plus 11 is 10 over 4 is 5/2. So these are my remaining zeros and my final list of zeros will be the original 2 that I had, 2 plus 3i, 2 minus -3i, -3 and 5/2.
Remember the Conjugate Zeros Theorem says that imaginary zeros come in conjugate pairs and the other theorem that we studied says that the number of zeros of a polynomial function equal its degree, this was a fourth degree polynomial function and we had four zeros.
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