what are the roots of x^3-3x^2+3x-1
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obviously i tried now help me cuz im confused
It's not at all obvious that you tried. I did the first one for you. Take a look at that one, and use the same method for the others. The chances are good that there will be at least one root that is easy to find.
At least show what ideas you tried, which failed.
ok thank you steve
im still confused
To find the roots of the polynomial function \(f(x) = x^3 - 3x^2 + 3x - 1\), we can use various methods such as factoring, the rational root theorem, or numerical methods. Let's start by factoring.
The given polynomial does not appear to have any obvious common factors between its coefficients. Therefore, we should consider other methods.
Next, we can check if the polynomial has any rational roots using the rational root theorem. According to the theorem, any rational root of \(f(x)\) will be of the form \(p/q\), where \(p\) is a factor of the constant term (-1) and \(q\) is a factor of the leading coefficient (1).
The factors of the constant term (-1) are ±1, and the factors of the leading coefficient (1) are ±1 as well. So, the possible rational roots are ±1.
We can substitute each possible root into \(f(x)\) and check if it yields 0. Evaluating \(f(x)\) at \(x = 1\), we get:
\(f(1) = 1^3 - 3(1)^2 + 3(1) - 1 = 0\)
Therefore, \(x = 1\) is a root of the given polynomial.
To further factorize the polynomial, we can divide it by the root we found (x - 1).
Using polynomial long division or synthetic division, we can divide \(f(x)\) by \(x - 1\) to obtain the quotient:
\(x^2 - 2x + 1\)
The quotient \(x^2 - 2x + 1\) is a quadratic polynomial, which we can further factorize using either factoring, completing the square, or using the quadratic formula.
In this case, \(x^2 - 2x + 1\) can be factored as \((x - 1)(x - 1)\).
So, the factored form of the polynomial is:
\(f(x) = (x - 1)(x - 1)(x - 1)\)
From this, we can see that the polynomial has a triple root of \(x = 1\).