How do you factor -3x^2+2x+2?

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Answer 1 · 2018-05-05T13:23:54

$- (3 x^{2} - 2 x - 2)$ $-(3x^2-2x-2)$

$- 3 x^{2} + 2 x + 2$ $-3x^2+2x+2$

Factor the negative out,

$- (3 x^{2} - 2 x - 2)$ $-(3x^2-2x-2)$

Answer 2 · 2018-05-05T13:25:56

$(x + \frac{1 - \sqrt{7}}{3}) (x + \frac{1 + \sqrt{7}}{3})$ $(x+(1-sqrt7)/3)(x+(1+sqrt7)/3)$

$- 3 x^{2} + 2 x + 2$ $-3x^2+2x+2$

Using the quadratic formula given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Since the equation is in the general form $a x^{2} + b x + c$ $ax^2+bx+c$ where $a = - 3$ $a=-3$ , $b = 2$ $b=2$ and $c = 2$ $c=2$

$x = \frac{- 2 \pm \sqrt{4 + 24}}{- 6}$ $x=(-2+-sqrt(4+24))/(-6$

$x = \frac{- 2 \pm \sqrt{28}}{- 6}$ $x=(-2+-sqrt28)/-6$

$x = \frac{- 2 \pm 2 \sqrt{7}}{- 6}$ $x=(-2+-2sqrt7)/-6$

$x = \frac{- 1 \pm \sqrt{7}}{- 3}$ $x=(-1+-sqrt7)/-3$

Therefore,

$(x - (\frac{- 1 + \sqrt{7}}{- 3})) (x - (\frac{- 1 - \sqrt{7}}{- 3}))$ $(x-((-1+sqrt7)/-3))(x-((-1-sqrt7)/-3))$

= $(x - (- 1) \frac{1 - \sqrt{7}}{3}) (x - (- 1) \frac{1 + \sqrt{7}}{3})$ $(x-(-1)(1-sqrt7)/3)(x-(-1)(1+sqrt7)/3)$

= $(x + \frac{1 - \sqrt{7}}{3}) (x + \frac{1 + \sqrt{7}}{3})$ $(x+(1-sqrt7)/3)(x+(1+sqrt7)/3)$