Find x-intercepts for... (1 Viewer)

Utility · Jun 7, 2012

y = x^3-2x^2-4x+8
0 = x^3-2x^2-4x+8

I got x = + / - 2, but I'd really like to see how someone else derives their answer. Thanks.

Carrotsticks · Jun 7, 2012

Utility said:
y = x^3-2x^2-4x+8
0 = x^3-2x^2-4x+8

I got x = + / - 2, but I'd really like to see how someone else derives their answer. Thanks.

By observation, I can see that you can easily factorise the cubic:

$y=x^3-2x^2-4x+8 = x^2(x-2) -4(x-2) = (x^2-4)(x-2) = (x-2)^2 (x+2)$

And so therefore, there is a single root at x=-2 and a DOUBLE root at x=2.

SpiralFlex1 · Jun 7, 2012

$$Factorise by grouping,$

$x^3-2x^2-4x+8=0$

$x^2(x-2)-4(x-2)=0$

$(x^2-4)(x-2)=0$

$(x-2)^2(x+2)=0$

$x=2,-2$

Find x-intercepts for... (1 Viewer)

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