Factorize $x^{3} - 7 x + 6$ $x^3-7x+6$ ?

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Factorize $x^{3} - 7 x + 6$ $x^3-7x+6$ ?

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Answer 1 · 2016-12-29T08:47:46

Factors of $x^{3} - 7 x + 6 = 0$ $x^3-7x+6=0$ are $(x - 1) (x + 2) (x - 3) = 0$ $(x-1)(x+2)(x-3)=0$

Explanation:

As the coefficients ( $1 - 7 + 6 = 0$ $1-7+6=0$ ) in the given equation add up to zero, it is apparent that $x = 1$ $x=1$ is a solution of $x^{3} - 7 x + 6 = 0$ $x^3-7x+6=0$ and $(x - 1)$ $(x-1)$ is a factor of $x^{3} - 7 x + 6$ $x^3-7x+6$ .

Dividing $x^{3} - 7 x + 6$ $x^3-7x+6$ by $(x - 1)$ $(x-1)$

$x^{3} - 7 x + 6 = x^{2} (x - 1) + x (x - 1) - 6 (x - 1)$ $x^3-7x+6=x^2(x-1)+x(x-1)-6(x-1)$

= $(x - 1) (x^{2} + x - 6)$ $(x-1)(x^2+x-6)$ - now splitting the middle term

= $(x - 1) (x^{2} + 2 x - 3 x - 6)$ $(x-1)(x^2+2x-3x-6)$

= $(x - 1) (x (x + 2) - 3 (x + 2))$ $(x-1)(x(x+2)-3(x+2))$

= $(x - 1) (x + 2) (x - 3)$ $(x-1)(x+2)(x-3)$

Hence factors of $x^{3} - 7 x + 6 = 0$ $x^3-7x+6=0$ are $(x - 1) (x + 2) (x - 3) = 0$ $(x-1)(x+2)(x-3)=0$

Answer 2 · 2016-12-29T08:47:48

The answer is $= (x - 1) (x + 3) (x - 2)$ $=(x-1)(x+3)(x-2)$

Explanation:

Let $f (x) = x^{3} - 7 x + 6$ $f(x)=x^3-7x+6$

$f (1) = 1 - 7 + 6 = 0$ $f(1)=1-7+6=0$

Therefore,

$(x - 1)$ $(x-1)$ is a factor

To find the other factors, we do a long division

$a a a a$ $color(white)(aaaa)$ $x^{3}$ $x^3$ $a a a a a$ $color(white)(aaaaa)$ $- 7 x + 6$ $-7x+6$ $a a a a a a$ $color(white)(aaaaaa)$ $∣$ $∣$ $x - 1$ $x-1$

$a a a a$ $color(white)(aaaa)$ $x^{3} - x^{2}$ $x^3-x^2$ $a a a a$ $color(white)(aaaa)$ #color(white)(aaaaaaaaaa)∣# $x^{2} + x - 6$ $x^2+x-6$

$a a a a a$ $color(white)(aaaaa)$ $0 + x^{2} - 7 x$ $0+x^2-7x$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ x^{2} - x$ $+x^2-x$

$a a a a a a a a$ $color(white)(aaaaaaaa)$ $+ 0 - 6 x + 6$ $+0-6x+6$

$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ $- 6 x + 6$ $-6x+6$

$a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $- 0 + 0$ $-0+0$

Therefore,

$\frac{x^{3} - 7 x + 6}{x - 1} = x^{2} + x - 6 = (x + 3) (x - 2)$ $(x^3-7x+6)/(x-1)=x^2+x-6=(x+3)(x-2)$

So,

$(x^{3} - 7 x + 6) = (x - 1) (x + 3) (x - 2)$ $(x^3-7x+6)=(x-1)(x+3)(x-2)$

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