How do you factor and find the zeroes for $F (x) = x^{3} + x^{2} + 4 x + 4$ $F(x) = x^3 + x^2 + 4x + 4$ ?

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Answer 1 · 2015-06-06T10:51:13

Factor by grouping...

$F (x) = x^{3} + x^{2} + 4 x + 4$ $F(x) = x^3+x^2+4x+4$

$= (x^{3} + x^{2}) + (4 x + 4)$ $= (x^3+x^2)+(4x+4)$

$= x^{2} (x + 1) + 4 (x + 1)$ $= x^2(x+1)+4(x+1)$

$= (x^{2} + 4) (x + 1)$ $= (x^2+4)(x+1)$

The $(x^{2} + 4)$ $(x^2+4)$ factor has no linear factors with real coefficients since $x^{2} + 4 \geq 4 > 0$ $x^2+4 >= 4 > 0$ for all $x in RR$ .

$F(x)=0$ has one real root, viz $x = -1$

How do you factor and find the zeroes for F(x) = x^3 + x^2 + 4x + 4F(x)=x3+x2+4x+4F(x) = x^3 + x^2 + 4x + 4?