Question 656796: Please help me solve in the real numbers system : x^3-3x^2+4=0
Found 2 solutions by htmentor, solver91311:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Please help me solve in the real numbers system : x^3-3x^2+4=0:
Using long division, you can verify that one of the factors is x+1, leaving the resulting equation:
x^2 - 4x + 4
This can be factored as
(x-2)(x-2)
So the equation fully factored is (x+1)(x-2)(x-2) = 0
So the solutions are x=-1, x=2 (multiplicity 2)
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The first step is to determine how many zeros you expect to find altogether. Since this is a third degree polynomial equation, the Fundamental Theorem of Algebra promises three zeros, counting all multiplicities.
Use the Rational Root Theorem. If the polynomial has rational zeros, then they will be of the form  where  is an integer factor of the constant term and  is an integer factor of the lead coefficient.
Your lead coefficient only has one integer factor, namely  . So the only possible rational roots all have a denominator of .
The integer factors of the constant term are ,  , and  . Note that these are not guaranteed to be zeros, in fact, there is no guarantee that there are any rational zeros. It is just that if a rational zero does exist, it will be in the set of numbers just described.
Use synthetic division to test the possible zeros one-by-one until you either find one that works or you have exhausted all of the possibilities and have proven thereby that there are no rational zeros. The latter case being another story for another time.
If you need a refresher on Synthetic Division, Click Here
1 | 1 -3 0 4
| 1 -2 -2
----------------
1 -2 -2 2
If you divide by  , then the last number in the bottom row of the synthetic division is the value of ) . Since \ \neq\ 0) , 1 is NOT a zero.
-1 | 1 -3 0 4
| -1 4 -4
----------------
1 -4 4 0
-1 IS a zero so  is a factor AND  is the other factor (note the coefficients come from the other numbers in the bottom row).
Since the remaining factor is a factorable quadratic, you no longer need to test for any other rational roots.
Factor the quadratic:
(x\ -\ 2))
Hence 2 is a zero with a multiplicity of 2.
The three zeros are then 2, 2, and -1.
John
My calculator said it, I believe it, that settles it
|
|
|
