SOLUTION: Find all zeros of the function f(x)=x^2(x-3)(x^3-8)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all zeros of the function f(x)=x^2(x-3)(x^3-8)      Log On


   



Question 825731: Find all zeros of the function f(x)=x^2(x-3)(x^3-8)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E2%28x-3%29%28x%5E3-8%29
Finding zeros of a polynomial usually involves factoring the polynomial. This function is already partially factored so some of our work has already been done.

The factor at the end, x%5E3-8, is a difference of cubes, x%5E3-2%5E3. So we can use the difference of cubes pattern, a%5E3-b%5E3=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29, to factor it:
f%28x%29=x%5E2%28x-3%29%28x-2%29%28x%5E2%2B2x%2B4%29
The last factor will not factor further. But it is a quadratic so we can use the quadratic formula to find its roots:
x+=+%28-%282%29+%2B-+sqrt%28%282%29%5E2-4%281%29%284%29%29%29%2F2%281%29
Simplifying...
x+=+%28-2+%2B-+sqrt%284-4%281%29%284%29%29%29%2F2
x+=+%28-2+%2B-+sqrt%284-16%29%29%2F2
x+=+%28-2+%2B-+sqrt%28-12%29%29%2F2
x+=+%28-2+%2B-+sqrt%28-1%2A4%2A3%29%29%2F2
x+=+%28-2+%2B-+sqrt%28-1%29%2Asqrt%284%29%2Asqrt%283%29%29%2F2
x+=+%28-2+%2B-+i%2A2%2Asqrt%283%29%29%2F2
x+=+%28-2+%2B-+2i%2Asqrt%283%29%29%2F2
x+=+%282%28-1+%2B-+i%2Asqrt%283%29%29%29%2F2
x+=+%28cross%282%29%28-1+%2B-+i%2Asqrt%283%29%29%29%2Fcross%282%29
x+=+-1+%2B-+i%2Asqrt%283%29
which is short for:
x+=+-1+%2B+i%2Asqrt%283%29 or x+=+-1+-+i%2Asqrt%283%29
These are two of the roots of f(x).

We will get the remaining roots from the other factors:
f%28x%29=x%5E2%28x-3%29%28x-2%29%28x%5E2%2B2x%2B4%29
The other roots will be the values for x that make a factor zero. For the first factor, x%5E2, we get a root of x = 0. And since x is a factor twice in x%5E2, zero counts as a root twice! (This is called a double root.)

For the other factors we should get roots of x = 3 and x = 2.

Altogether, the roots of f(x) are: 0, 0, 2, 3, -1+%2B+i%2Asqrt%283%29 and -1+-+i%2Asqrt%283%29

P.S. This is a polynomial, not a rational function. Please post in an appropriate category.