SOLUTION: This is from a study guide for my final the question is: Find all the real and complex zeros of the polynomial x^4+4x^3-4x^2-36x-45 ...I know the answe involves 3, 2 and an ima

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: This is from a study guide for my final the question is: Find all the real and complex zeros of the polynomial x^4+4x^3-4x^2-36x-45 ...I know the answe involves 3, 2 and an ima      Log On


   

Question 41075: This is from a study guide for my final the question is:
Find all the real and complex zeros of the polynomial x^4+4x^3-4x^2-36x-45
...I know the answe involves 3, 2 and an imaginary number but I am having difficulty which ones are postive and negative etc...
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the real and complex zeros of the polynomial x^4+4x^3-4x^2-36x-45
The polynomial facors as (x-3)(x+3)(x^2+4x+5)
The zeroes are 3, -3 and the zeroes of x^2+4x+5 which are
worked out below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A1%2A5=-4.

The discriminant -4 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -4 is + or - sqrt%28+4%29+=+2.

The solution is x%5B12%5D+=+%28-4%2B-+i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28-4%2B-+i%2A2%29%2F2%5C1+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B5+%29

Hope this helps.
Cheers,
Stan H.