Math Question - Real and Complex Zeros?

Question

I'm completely lost in my precalculus now. There was no school today and I have a test tomorrow and I was hoping that someone could explain how to find the real and complex zeros of the polynomial function below.

y = x^4 - 2x^3 + x^2 - 8x - 12

So far, I have found the real zeros, which are x=3, and x= -1. From what I can remember, the degree of the function tells you how many zeros there are, so wouldn't there be two more zeros that I need to find?

Answer 1

Do you know how to do synthetic division?

An easy way to start is to graph this in the calculator and then look at teh table to see what x values on the graph make y=0. At x=-1 and x=3 y=0 so these are your two real zeros.

If x=-1 then it is (x+1) in the equation and if x=3 then it equals (x-3) in the equation.

Using either synthetic or long division you can then find that x^2+4 is the unnacounted portion of your equation.

If x^2 +4=0, then x^2= -4 Taking the square root of both sides yields x= +2i or -2i.

Your equation has real zeros at x=-1 and x=3 and imaginary solutions (complex) at x=-2i and x=2i

I hope this will help some.

Answer 2

Let y = P(x) = x^4 -2x^3 + x^2 -8x -12
To find if 1 is a zero, put x = 1
P (1) = 1^4 -2 (1^3) + 1^2 - 8 (1) -12 = 1 -2 + 1 - 8 -12 = -20
P(-1) = 1 +2 + 1 + 8 -12 = 0
Hence -1 is one zero. x = -1 means x+1 = 0
Dividing x^4 -2x^3 + x^2 -8x -12 by x+1 we get
x^3 - 3x^2 + 4x -12
You could proceed in the same manner as above finding P(-1), P(2), P(-2) etc
But on observation we get
x^2(x-3) + 4(x-3)
(x-3)(x^2 +4)
x-3 = 0; x =3 is the next zero
x^2 + 4 = x^2 - 4 i^2 = (x + 2i)(x-2i)
x+2i =0; x=-2i
x-2i = 0; x = 2i
The real zeroes are -1 &3
The complex zeroes are 2i or -2i

All the best in tomorrow's test

Answer 3

Finding Complex Zeros

Answer 4

First of all you need to find possible roots that are all factors of -12: +1, -1, +2, -2, +3, -3, +4, -4, +6, -6, +12 , -12

By using reminder theorem you substitute every single possible factor in your equationa, and where you find a zero, that will be a factor.

example:
for x = 1

1 ^4 -2*1^3 +1^2 -8*1 -12 different of zero so x = 1 is not a factor

you can also find synthetic division of the polynomial expression divided by x-1 and if it has no reminder it is a factor.

Answer 5

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RE:
Math Question - Real and Complex Zeros?
I'm completely lost in my precalculus now. There was no school today and I have a test tomorrow and I was hoping that someone could explain how to find the real and complex zeros of the polynomial function below.

y = x^4 - 2x^3 + x^2 - 8x - 12

Answer 6

Yes, there are 2 more zeros you need to find.
When you divide out x-3 and x+1 you
get x²+4. So the other 2 zeros are 2i and -2i.

Answer 7

g(x) = x^4 + 2x^3 + x^2 + 8x - 12 => g(x) = 0 => x^4 + 2x^3 + x^2 + 8x - 12 = 0 => x^4 - x^3 + 3x^3 - 3x^2 + 4x^2 - 4x + 12x - 12 = 0 => x^3 (x - 1) + 3x^2 (x - 1) + 4x (x - 1) + 12 (x - 1) = 0 => (x - 1) (x^3 + 3x^2 + 4x + 12) = 0 => (x - 1) [x^2(x + 3) + 4(x + 3)] = 0 => (x - 1) (x + 3) (x^2 + 4) = 0 => x = 1 and x = -3 are the real zeroes of g(x).

Answer 8

use synthetic division to verify the one of the zeros...then use synthetic division again on the 3rd degee polynominal that you are left with...one you have divided you will have a quadratic equation...either factor that or use the quadratic formula and that will give you the other 2 imaginary zeros...


or you you could foil out the 2 real zeros and use long division...either way will get the imaginary zeros...you have a question ask me