How do you factor and find the zeroes of the polynomial $y = - 2 x^{4} + 10 x^{3} - 12 x^{2}$ $y=-2x^4+10x^3-12x^2$ ?

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Answer 1 · 2017-02-08T22:55:43

Zeroes at $x = 3, x = 2 and x = 0$ $x= 3, x= 2 and x= 0$ .

Factor out an $- 2 x^{2}$ $-2x^2$ :

$y = - 2 x^{2} (x^{2} - 5 x + 6)$ $y = -2x^2(x^2 - 5x + 6)$

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

Set to $0$ $0$ and solve for $x$ $x$ :

$0 = - 2 x^{2} (x - 3) (x - 2)$ $0 = -2x^2(x - 3)(x- 2)$

$x = 0, 3 and 2$ $x = 0, 3 and 2$

This has zeroes at $x = 3, x = 2 and x = 0$ $x = 3, x = 2 and x = 0$ .

Hopefully this helps!

How do you factor and find the zeroes of the polynomial y=-2x^4+10x^3-12x^2y=−2x4+10x3−12x2y=-2x^4+10x^3-12x^2?