To find the zeros of the polynomial function f(x) = x³ + x² – 20x, we need to solve the equation f(x) = 0. This means we want to find the values of x for which the function equals zero.
First, we start by factoring the polynomial. We can rearrange the function as follows:
f(x) = x(x² + x – 20)
Next, we can see that one factor is x, which gives us the first zero:
x = 0
Now, we need to solve the quadratic equation x² + x – 20 = 0. We can do this by using the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a
In this case, a = 1, b = 1, and c = -20. Plugging these values into the quadratic formula gives us:
x = (-1 ± √(1² – 4(1)(-20))) / (2(1))
This simplifies to:
x = (-1 ± √(1 + 80)) / 2
x = (-1 ± √81) / 2
x = (-1 ± 9) / 2
Now we have two solutions:
- x = (8) / 2 = 4
- x = (-10) / 2 = -5
Therefore, we find that the zeros of the polynomial function f(x) = x³ + x² – 20x are:
- x = 0
- x = 4
- x = -5
In conclusion, the complete set of zeros for the polynomial is 0, 4, and -5.