x³ + 10x² + 24x = 0
Next, we can factor out the common term from each part of the polynomial. In this case, we can factor out an x:
x(x² + 10x + 24) = 0
Now, we can see that one zero is x = 0. To find the other zeros, we need to solve the quadratic equation x² + 10x + 24 = 0.
We can apply the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a
Here, a = 1, b = 10, and c = 24. Plugging in these values gives us:
x = (−10 ± √(10² − 4(1)(24))) / (2(1))
x = (−10 ± √(100 − 96)) / 2
x = (−10 ± √4) / 2
x = (−10 ± 2) / 2
This results in two potential solutions:
x = (−10 + 2) / 2 = −8 / 2 = −4
and
x = (−10 – 2) / 2 = −12 / 2 = −6
Thus, the zeros of the polynomial function f(x) = x³ + 10x² + 24x are:
- x = 0
- x = −4
- x = −6