what are the roots of this equation x^3-3x+2=0?

Answer 1

You can see that 1 is a root of this equation.
So, x-1 is a factor.
Divide x³-3x+2 by x-1 to get
x²+x-2.
The latter factors into (x+2)(x-1)
So your equation can be written
(x-1)(x-1)(x+2) = 0,
and the roots are x = 1(double root) and x = -2.

Just for fun, try x^4-4x+3 and x^5-5x+4
and see what you get for factors. An interesting
pattern will emerge!

Answer 2

It's a depressed cubic. Solution given by Scipione dal Ferro (early16th century).

Rewrite in the form x^3 + A x = B

In this case: A = -3 and B = -2

Next, find two numbers, called s and t, such that
3st = A and
s^3 - t^3 = B

Then x = s-t is a root.

s=-1 and t = 1
3st = 3*(-1)*1 = -3 = A
(-1)^3 - (1)^3 = -1 -1 = -2 = B

x = -1 -(1) = -2 should be a root.

Check

(-2)^3 - 3(-2) +2 = -8 + 6 + 2 = 0
OK

divide x^3-3x+2 by (x+2) to get:
x^2 - 2x + 1
which is a quadratic (much easier, go for it)

Answer 3

1 and -2

Answer 4

x^3-3x+2=0
x=1 is an obvious solution so (x-1) must be a factor
Dividing x^3-3x+2 by x-1 gives x^2+x-2 = (x+2)(x-1)
So x^3-3x+2 = (x-1)(x-1)(x+2)
The roots are 1,1,-2

Answer 5

You can use the Quadratic Formula to solve this

I would have to verbalize the equation since there is no way to type it on computer


x equals [negative b plus or minus (Sign) square root of (b squared minus 4 times a times c)] over two a

When you perform the necessary steps, you get

x= [3 plus or minus square root of 1] over 2

Answer 6

1 and -2 x^3-3x+2=(x-1)^2 ( x+2)

Answer 7

(x+2)(x-1)^2
x=-2, x=1.

Answer 8

mmmm