Question 1011575
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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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{{{x^4+2x^3+x^2+8x-12}}} = {{{0}}}.


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 {{{x^4+2x^3+x^2+8x-12}}} by (x-1). You will get 

{{{(x^4+2x^3+x^2+8x-12)/(x-1)}}} = {{{x^3 + 3x^2 + 4x + 12}}}.

The polynomial {{{x^3 + 3x^2 + 4x + 12}}} 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.
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