The problem is y= x(x-1)^3

There are many ways to find. First of all, if a function is in factored form, you can set the factors equal to 0 and solve them one at a time. In this case we have x = 0 or x - 1 = 0. Solving gives x = 0 or 1.

For functions where it is not obvious or impossible to find closed form expressions for the zeros, there are methods for finding numerical approximations for the zeros with arbitrary precision. Here are a few, each with it's own advantages and disadvantages.

1. Bisection method.

Assume that f(x) is continuous in the interval [a,b] and that f(a) and f(b) have opposite signs.

Repeat the following steps until desired precision is attained:

A. Let c = (a + b) / 2 be the midpoint,

B. If the sign of f(a) and f(c) are opposite, then there is a root in [a,c] so set b equal to c.

C. Otherwise, the sign of f(b) and f(c) are opposite, so there is a root in [c,b] so set a equal to c.

2. Secant method.

Start with two guesses, x_{0} and x_{1}.

For n = 0,1,2..., let x_{n+2} be the x-intercept of the line through (x_{n+1}, f(x_{n+1})) and (x_{n}, f(x_{n})).

Assuming the function is continuous, and the sequence converges (has a limit), then the limit is a root.

3. Newton's method

Start with a guess x_{0}. Then for n=0,1,2,..., let x_{n+1} be the x-intercept of the line tangent to f(x) at x=x_{n}.

The formula for such an x_{n+1} is x_{n+1} = x_{n} - f(x_{n}) / f'(x_{n}), where f' is the derivative (slope) of f.

If the function is continuous and the sequence converges, then the limit is a root.

You will learn this method when you take calculus.

Now for multiplicity of a root of any function.

The multiplicity of the root r of a function f(x) is the integer n such that r is a root of f, f', f'', ..., f^{(n-1)} but not a root of f^{(n)}. This definition uses calculus.

For example, with your y = x(x-1)^3 = x^{4} - 3x^{3} + 3x^{2} - x, we have.

y' = 4x^{3} - 9x^{2} + 6x - 1

y'' = 12x^{2} - 18x^{ }+ 6

y''' = 24x - 18

Since at x = 0, we have y = 0 but y' ≠ 0, the multiplicity of x = 0 is 1.

Since at x = 1, we have y = y' = y'' = 0, but y''' ≠ 0, the multiplicity of x = 1 is 3.

In the case of a polynomial, the multiplicity of a root r is the number of times that you can factor out x - r.

Using this we find that the the multiplicity of x = 0 is 1, and the multiplicity of x = 1 is 3, which agrees with the above result.