What are the zeros of the function? What are their multiplicities?
f(x) = 3x3 + 15x2 + 18x
work out
To find the zeros of the function f(x), we need to set f(x) equal to zero and solve for x:
3x^3 + 15x^2 + 18x = 0
Factoring out an x from each term:
x(3x^2 + 15x + 18) = 0
Setting each factor equal to zero:
x = 0
3x^2 + 15x + 18 = 0
Now, we can solve the quadratic equation for the remaining zeros:
Using the quadratic formula, where ax^2 + bx + c = 0:
x = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation 3x^2 + 15x + 18 = 0:
a = 3, b = 15, c = 18
Plugging in these values, the quadratic formula becomes:
x = (-15 ± √(15^2 - 4(3)(18))) / (2(3))
x = (-15 ± √(225 - 216)) / 6
x = (-15 ± √9) / 6
x = (-15 ± 3) / 6
x = (-15 + 3) / 6 or x = (-15 - 3) / 6
x = -12/6 or x = -18/6
Simplifying:
x = -2 or x = -3
Therefore, the zeros of the function f(x) = 3x^3 + 15x^2 + 18x are x = 0, x = -2, and x = -3.
To determine the multiplicities of the zeros, we can look at the exponents of the factors in the factored form of the function.
The factor x has a multiplicity of 1 since it is raised to the first power.
The factor (3x^2 + 15x + 18) can be factored further:
3x^2 + 15x + 18 = 3(x^2 + 5x + 6) = 3(x + 2)(x + 3)
The factors (x + 2) and (x + 3) are each raised to the first power, indicating they have a multiplicity of 1.
Therefore, the multiplicities of the zeros of f(x) are as follows:
Multiplicity of 0: 1
Multiplicity of -2: 1
Multiplicity of -3: 1