How do you find the roots of 0=x^3 -3*x^2 +2?

Answer 1

The way to find the rational roots of:
a_n*x^n + a_n1 * x^(n-1) + ... + a_1 * x + a_0
is to look at all the possible factors of a_0 and divide them by all the possible factors of a_n (plus or minus) and those are your possibilities. This works in general. But it does not tell you the irrational or imaginary roots (those with a square root of a non square or negative number, for example).

In this case a_0 (the constant term) is 2
and a_n (the coefficient on the highest term) is 1
So you have to try plus/minus 1 and plus/minus 2
divided by 1.

Well, that's just +/-1 and +/-2
If any of those work, say x=r gives you a 0, then factor (x-r) out of the original and you've got a reduced polynomial.

Hint: exactly one of the numbers above does indeed produce a 0.

Answer 2

First look at the absolute coefficients of the terms - they go (1, 3, 2). Surely there is a simple value of x which will let the 1 and 2 take the opposite sign from the 3, so that everything cancels out? Yes, x = 1 turns the right-hand side into 1 - 3 + 2 which is zero, so 1 is a root, and (x - 1) is a factor.

Synthetic division turns the equation into 0 = (x - 1)*(x^2 - 2x - 2).

The roots of x^2 - 2x - 2 are not whole numbers, but with the quadratic formula you can find that they are (1 + sqrt(3)) and (1 - sqrt(3)).

Answer 3

Just try a few numbers. If you try 1 you get 0=1-3+2 =0.
So 1 is a root and x-1 is a factor.
Now use synthetic division or long division to find
(x^3-3x^2+2) divided by (x-1).

The result will be a quadratic equation which you can factor or use the quadratic formula to find the other two roots.

Piece of cake.

Answer 4

by inspection x = 1 is a root of the equation Thus x^3 -3*x^2 +2 = (x - 1)( x^2 -2x -2) etc

Answer 5

Graphing, synthetic division, or factoring.

I tried it out. The roots I got were 1 and 1 (plus or minus) radical 3.

x-1, x ± √3