Question 1127897
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You have the correct answer. It is choice A. Nice work.


To prove that (x+3) is a factor, plug x = -3 into f(x) and you should get 0 as a result. This is using the remainder theorem.


If you were to divide f(x) over (x+3), you should get x^2+2x-2 as the quotient. Solve x^2+2x-2=0 through the quadratic formula and you should get the following:
{{{x = (-b+sqrt(b^2-4ac))/(2a)}}} or {{{x = (-b-sqrt(b^2-4ac))/(2a)}}}
{{{x = (-2+sqrt((2)^2-4(1)(-2)))/(2(1))}}} or {{{x = (-2-sqrt((2)^2-4(1)(-2)))/(2(1))}}}
{{{x = (-2+sqrt(12))/(2)}}} or {{{x = (-2-sqrt(12))/(2)}}}
{{{x = (-2+sqrt(4*3))/(2)}}} or {{{x = (-2-sqrt(4*3))/(2)}}}
{{{x = (-2+sqrt(4)*sqrt(3))/(2)}}} or {{{x = (-2-sqrt(4)*sqrt(3))/(2)}}}
{{{x = (-2+2*sqrt(3))/(2)}}} or {{{x = (-2-2*sqrt(3))/(2)}}}
{{{x = -1+sqrt(3)}}} or {{{x = -1-sqrt(3)}}}
which help set up the other factor of the answer


Note: if x = k is a root, then x-k is a factor
So if {{{x = -1+sqrt(3)}}} is a root, then {{{x -( -1+sqrt(3))}}} is a factor
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