To find the roots of the function f(x) = x3 + x2 – 6x, we need to set the function equal to zero:
x3 + x2 – 6x = 0
The first step is to factor out the common term. Notice that x is a common factor:
x(x2 + x – 6) = 0
Now, we have one root as:
- x = 0
Next, we need to factor the quadratic equation x2 + x – 6. To do this, we look for two numbers that multiply to -6 (the constant term) and add up to 1 (the coefficient of x). These two numbers are 3 and -2.
Thus, we can factor the quadratic as:
(x + 3)(x – 2) = 0
Combining this with our earlier factored form, we have:
x(x + 3)(x – 2) = 0
Setting each factor equal to zero gives us the possible solutions (roots):
- x = 0
- x + 3 = 0 → x = -3
- x – 2 = 0 → x = 2
Therefore, the complete set of roots for the function f(x) = x3 + x2 – 6x is:
- x = 0
- x = -3
- x = 2
This means that the function intersects the x-axis at these points, and you can verify these roots by substituting back into the original function to ensure that each makes the function equal to zero.