What is the solution set for x^3-5x^2-4x+20=0 ?

The solution set is:
x3+5x2−4x−20=0Factor the left side of the equation.Factor out the greatest common factor from each group.Group the first two terms and the last two terms.(x3+5x2)+(−4x−20)=0Factor out the greatest common factor (GCF) from each group.x2(x+5)−4(x+5)=0Factor the polynomial by factoring out the greatest common factor, x+5.(x+5)(x2−4)=0Rewrite 4 as 22.(x+5)(x2−22)=0

Factor.Since both terms are perfect squares, factor using the difference of squares formula, a2−b2=(a+b)(a−b) where a=x and b=2.(x+5)((x+2)(x−2))=0Remove unnecessary parentheses.(x+5)(x+2)(x−2)=0
Set x+5 equal to 0 and solve for x.
Set the factor equal to 0.x+5=0Since 5 does not contain the variable to solve for, move it to the right side of the equation by subtracting 5 from both sides.x=−5
Set x+2 equal to 0 and solve for x.
Set the factor equal to 0.x+2=0Since 2 does not contain the variable to solve for, move it to the right side of the equation by subtracting 2 from both sides.x=−2
Set x−2 equal to 0 and solve for x.Set the factor equal to 0.x−2=0Since −2 does not contain the variable to solve for, move it to the right side of the equation by adding 2 to both sides.x=2
This means the solution is x+5=0,x+2=0, and x−2=0.x=−5,−2,2
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