SOLUTION: f(x)=x^4+2x^3+x^2+8x-12 I need to find all of the actual zeros of this function, including the complex. Thanks.

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Question 1011575: f(x)=x^4+2x^3+x^2+8x-12

I need to find all of the actual zeros of this function, including the complex. Thanks.
Found 4 solutions by ikleyn, Boreal, MathLover1, MathTherapy:
Answer by ikleyn(52887)   (Show Source):
You can put this solution on YOUR website!
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f(x)=x^4+2x^3+x^2+8x-12
I need to find all of the actual zeros of this function, including the complex. Thanks.
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= . It is easy to check that x = 1 is the root. It means that the left side polynomial has the factor (x-1). Make long division of by (x-1). You will get = . The polynomial has the root x = -3. (Check it). Then this polynomial has the factor (x+3). Make the long division again and get the quotient, which is the quadratic polynomial this time. I hope that you can complete yourself the assignment from this point.

Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website!
One can guess, using possible roots of +/-1,+/-2, +/-3, +/-4, +/-6, and +/-12.
The latter two are unlikely.
try
1:1//2//1//8//-12
=1=3=4=12=/0
(x-1) is a factor
What is left is x^3+3x^2+4x+12
-3:1//3//4//12
==1/0/4//0
(x+3) is a factor
left is x^2+4, which factors into (x+2i) and (x-2i)
The zeros are 1,-3, +2i,-2i

Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
....to factor it, first write as and as
group

solutions:
if =>=>
if =>
if =>

Answer by MathTherapy(10557)   (Show Source):
You can put this solution on YOUR website!
f(x)=x^4+2x^3+x^2+8x-12

I need to find all of the actual zeros of this function, including the complex. Thanks.

Using the rational root theorem, you'll find that 1 and - 3 are 2 of the roots.
Therefore, 2 of its factors are: (x - 1) and (x + 3), which combine to give the factor:
Dividing by the factor: () by long division or synthetic division results in the other factor:
= . Thus, the other 2 roots are:
Therefore, roots, or zeroes are: