It can have 1, 2 or 3 unique roots.
According to the rational root theorem, which of the following are possible roots of the polynomial function below?F(x) = 8x3 - 3x2 + 5x+ 15
A third degree polynomial could have one or three real roots.
4, the same as the degree of the polynomial.
Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)
A third-degree equation has, at most, three roots. A fourth-degree polynomial has, at most, four roots. APEX 2021
Each distinct real root is an x-intercept. So the answer is 4.
1
Four.Four.Four.Four.
5, Using complex numbers you will always get 5 roots.
NONE If both roots are imaginary, the means the parabola does NOT cross the x-axis at all. The place where a function crosses the x- axis has the coordinate (x,0) for some value of x. That means if you plug in x to the function or polynomial, you get 0. This is equivalent to saying that x is a root of the polynomial. But if the only roots are imaginary, there will be no point (x,0) for any real number x.
None, it involves the square root of a negative number so the roots are imaginary.