Question 1050510
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Find all real and complex zeros of the polynomial, and write p(x) as the product of linear factors. 
P(x)=x^4-9x^2+4x+12
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Apply grouping to factor P(x):

{{{x^4-9x^2+4x+12}}} = {{{(x^4-9x^2)+(4x+12)}}} = {{{x^2(x^2-9) + 4*(x+3)}}} = {{{x^2*(x+3)*(x-3) + 4*(x+3)}}} = {{{(x+3)*(x^2*(x-3)+4)}}} = {{{(x+3)*(x^3-3x^2+4)}}}. 

So, one root is x = -3.

Next, transform and factor the polynomial Q(x) = x^3 - 3x^2 + 4.
Add and distract x^2 and then apply grouping, again:

{{{x^3 - 3x^2 + 4}}} = {{{(x^3 + x^2) - (4x^2 -4)}}} = {{{x^2*(x+1) - 4*(x^2-1)}}} = {{{x^2*(x+1) - 4*(x+1)*(x-1)}}} = {{{(x+1)*(x^2-4)}}} = {{{(x+1)*(x+2)*(x-2)}}}.

The roots of the polynomial Q(x) are -1, -2, 2.

<U>Answer</U>. The roots of the polynomial P(x) are -3, -2, -1, and 2.
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