How many complex zeros does the polynomial function have? you.! f(x)=x^3-96x^2+400

the degree of that polynomial is 3, so by the fundamental theorem of algebra, it can have at most 3 solutions/zeros.

now, complex solutions, namely a + bi type, never come all by their lonesome, their sister is always with them, the conjugate, namely a - bi.

so if the equation does have any complex solutions/zeros, and it can have at most 3 zeros, then the complex ones must be just 2, since they always come in pairs.

None

Step-by-step explanation:

A graphing calculator shows you the function has 3 real zeros.

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You can guess that it will have all real zeros based on the coefficients. The constant of 400 tells you the y-intercept. You can see that for small values of x greater than 2 that the x^2 term can have a larger magnitude than the sum of the other two terms, so the function will be negative.

That means there are 3 real roots.

A cubic function (or any odd-degree function) always crosses the x-axis. Since the function tends toward -∞ for large negative values of x, there must be one zero-crossing between x = -∞ and x = 0, where the value is 400.

For values of x between about 3 and 95, f(x) will be negative again. Since the function ends up positive for large positive values of x, this means there must be two more zero crossings for positive values of x. There can only be 3 roots, so if all 3 are real, none are complex.

sorry, i don't know i just need points, i'm new here. chao

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