Factor $x^3 – 6x^2 + 5x + 12$ ?

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Answer 1 · 2018-02-23T03:43:50

$=> (x+1)(x-3)(x-4)$

Factors are $x = -1, 3, 4$

$(x^3 -6x^2 + 5x + 12)$

Since the sum of the coefficients of the terms is zero, $x +1$ is a factor.

Using synthetic division,

$color(white)(aaa) -1 color(white)(aaa)|color(white)(aaa)1color(white)(aaa)-6color(white)(aaa)5color(white)(aaa)12$

$color(white)(aaaaaaaaa)|color(white)(aaa)darrcolor(white)(aa)-1color(white)(aaa)7color(white)(a)-12$

$color(white)(aaaaaaaaa)|………………………$

$color(white)(aaaaaaaaaaaaaa)1color(white)(aaa)-7color(white)(aaa)12color(white)(aaa)0$

$=> (x+1) (x^2 - 7x + 12)$

$=> (x+1)(x^2 - 3x - 4x + 12)$

$=> (x + 1) (x(x - 3) -4 (x-3))$

$=> (x+1)(x-3)(x-4)$

Factors are $x = -1, 3, 4$

Factor x^3 – 6x^2 + 5x + 12x^3 – 6x^2 + 5x + 12 ?