Question 825731
{{{f(x)=x^2(x-3)(x^3-8)}}}
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.<br>
The factor at the end, {{{x^3-8}}}, is a difference of cubes, {{{x^3-2^3}}}. So we can use the difference of cubes pattern, {{{a^3-b^3=(a-b)(a^2+ab+b^2)}}}, to factor it:
{{{f(x)=x^2(x-3)(x-2)(x^2+2x+4)}}}
The last factor will not factor further. But it is a quadratic so we can use the quadratic formula to find its roots:
{{{x = (-(2) +- sqrt((2)^2-4(1)(4)))/2(1)}}}
Simplifying...
{{{x = (-2 +- sqrt(4-4(1)(4)))/2}}}
{{{x = (-2 +- sqrt(4-16))/2}}}
{{{x = (-2 +- sqrt(-12))/2}}}
{{{x = (-2 +- sqrt(-1*4*3))/2}}}
{{{x = (-2 +- sqrt(-1)*sqrt(4)*sqrt(3))/2}}}
{{{x = (-2 +- i*2*sqrt(3))/2}}}
{{{x = (-2 +- 2i*sqrt(3))/2}}}
{{{x = (2(-1 +- i*sqrt(3)))/2}}}
{{{x = (cross(2)(-1 +- i*sqrt(3)))/cross(2)}}}
{{{x = -1 +- i*sqrt(3)}}}
which is short for:
{{{x = -1 + i*sqrt(3)}}} or {{{x = -1 - i*sqrt(3)}}}
These are two of the roots of f(x).<br>
We will get the remaining roots from the other factors:
{{{f(x)=x^2(x-3)(x-2)(x^2+2x+4)}}}
The other roots will be the values for x that make a factor zero. For the first factor, {{{x^2}}}, we get a root of x = 0. And since x is a factor twice in {{{x^2}}}, zero counts as a root twice! (This is called a double root.)<br>
For the other factors we should get roots of x = 3 and x = 2.<br>
Altogether, the roots of f(x) are: 0, 0, 2, 3, {{{-1 + i*sqrt(3)}}} and {{{-1 - i*sqrt(3)}}}<br>
P.S. This is a polynomial, not a rational function. Please post in an appropriate category.