Answer :

RSTalbut
First get all the x terms on the left, so you get:
x³-6x²+8x=0
Then I used long division of polynomials and trial and error to factorise it:
(x³-6x²+8x)÷(x-2)=(x²-4x)
So:
(x³-6x²+8x)=(x²-4x)(x-2)=x(x-4)(x-2)
Therefore, for this cubic to equal 0, x=0, 4, or 2. 
I hope this is clear enough to help! :)
AL2006
x³ + 8x  =  6x²

Subtract  6x²  from each side:

x³ - 6x² + 8x  =  0

Before you go any further, factor 'x' out of the left side
and then let's talk about it:

x (x² - 6x + 8)  =  0

This equation is true when  x = 0 , so that's one of the solutions.

The other two will emerge when you figure out what makes (x² - 6x + 8) = 0.

Can that be factored ?  How about  (x - 4) (x - 2) ?

Now we can say that we need  (x - 4) (x - 2) = 0 .
That's true when  x = 4  or x = 2 .

So the three values of  'x'  that make the original equation a
true statement are . . .

x = 0
x = 2
x = 4

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