Given that the sum of two of the roots of $x^{3} - 3 x^{2} - 4 x + 12 = 0$ $x^3-3x^2-4x+12=0$ is zero, how do you factor $x^{3} - 3 x^{2} - 4 x + 12$ $x^3-3x^2-4x+12$ ?

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Given that the sum of two of the roots of $x^{3} - 3 x^{2} - 4 x + 12 = 0$ $x^3-3x^2-4x+12=0$ is zero, how do you factor $x^{3} - 3 x^{2} - 4 x + 12$ $x^3-3x^2-4x+12$ ?

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Answer 1 · 2017-03-12T12:01:29

$x^{3} - 3 x^{2} - 4 x + 12 = (x - 2) (x + 2) (x - 3)$ $x^3-3x^2-4x+12 = (x-2)(x+2)(x-3)$

Explanation:

If the roots are $α$ $alpha$ , $β$ $beta$ and $γ$ $gamma$ then:

$x^{3} - 3 x^{2} - 4 x + 12 = (x - α) (x - β) (x - γ)$ $x^3-3x^2-4x+12 = (x-alpha)(x-beta)(x-gamma)$

$x^{3} - 3 x^{2} - 4 x + 12 = x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ$ $color(white)(x^3-3x^2-4x+12) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma$

So, equating the coefficients of $x^{2}$ $x^2$ , we find:

$α + β + γ = 3$ $alpha+beta+gamma = 3$

Since the sum of two of the roots is $0$ $0$ , the remaining one must be $3$ $3$ .

So $(x - 3)$ $(x-3)$ is a factor.

In fact we find:

$x^{3} - 3 x^{2} - 4 x + 12 = x^{2} (x - 3) - 4 (x - 3)$ $x^3-3x^2-4x+12 = x^2(x-3)-4(x-3)$

$x^{3} - 3 x^{2} - 4 x + 12 = (x^{2} - 4) (x - 3)$ $color(white)(x^3-3x^2-4x+12) = (x^2-4)(x-3)$

$x^{3} - 3 x^{2} - 4 x + 12 = (x^{2} - 2^{2}) (x - 3)$ $color(white)(x^3-3x^2-4x+12) = (x^2-2^2)(x-3)$

$x^{3} - 3 x^{2} - 4 x + 12 = (x - 2) (x + 2) (x - 3)$ $color(white)(x^3-3x^2-4x+12) = (x-2)(x+2)(x-3)$

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Given that the sum of two of the roots of $x^{3} - 3 x^{2} - 4 x + 12 = 0$ $x^3-3x^2-4x+12=0$ is zero, how do you factor $x^{3} - 3 x^{2} - 4 x + 12$ $x^3-3x^2-4x+12$ ?

1 Answer

Explanation:

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Given that the sum of two of the roots of x3−3x2−4x+12=0x^3-3x^2-4x+12=0 is zero, how do you factor x3−3x2−4x+12x^3-3x^2-4x+12 ?

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Explanation:

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Given that the sum of two of the roots of $x^{3} - 3 x^{2} - 4 x + 12 = 0$ $x^3-3x^2-4x+12=0$ is zero, how do you factor $x^{3} - 3 x^{2} - 4 x + 12$ $x^3-3x^2-4x+12$ ?