what is an example or a quintic function with exactly four real zeros?
y = x^2(x-1)(x-2)(x-3)
notice the "double root" of x = 0
To find an example of a quintic function with exactly four real zeros, let's start by considering the general form of a quintic function:
f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f
To have exactly four real zeros, the function needs to have one double root and two distinct real roots. Let's suppose the double root is "r" and the two distinct real roots are "s" and "t".
Using these assumptions, we can expand the function as follows:
f(x) = (x - r)^2 * (x - s) * (x - t)^2
Expanding this expression will give us a quintic function with four real zeros. However, keep in mind that there are infinitely many possibilities for the values of "a", "b", "c", "d", "e", and "f" that will yield four real zeros.
To find an example of a quintic function with exactly four real zeros, we need to create a function of degree 5 that has four distinct real solutions.
One way to do this is by starting with a basic quintic function that has five real zeros, and then removing one of the zeros by making it a double root.
For example, let's start with the function f(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5). This function has five real zeros: x = 1, 2, 3, 4, and 5.
To make it have exactly four real zeros, we can change one of the zeros into a double root by squaring one of the factors. Let's square the factor (x - 3), so the new function becomes:
g(x) = (x - 1)(x - 2)(x - 3)^2(x - 4)(x - 5).
Now, the zero x = 3 has become a double root because of the squared factor (x - 3)^2. The other four zeros x = 1, 2, 4, and 5 remain as single roots.
Therefore, the example of a quintic function with exactly four real zeros is g(x) = (x - 1)(x - 2)(x - 3)^2(x - 4)(x - 5).