Answer:
B. 0 with multiplicity 3, 3 with multiplicity 2, and 2 with multiplicity 1
Step-by-step explanation:
Took a test
Factoring and solving the equation, it is found that the zeroes are:
B. 0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1.
The function is:
[tex]f(x) = -x^5 + 9x^4 - 18x^3[/tex]
Simplifying by [tex]-x^3[/tex]:
[tex]f(x) = -x^3(x^2 - 9x + 18)[/tex]
The zeroes are the values of x for which:
[tex]f(x) = 0[/tex]
[tex]-x^3 = 0 \rightarrow x = 0[/tex]
0 has multiplicity 3.
And:
[tex]x^2 - 9x + 18 = 0[/tex]
Quadratic equation with [tex]a = 1, b = -9, c = 18[/tex].
[tex]\Delta = (-9)^{2} - 4(1)(18) = 9[/tex]
[tex]x_{1} = \frac{-(-9) + \sqrt{9}}{2(1)} = 6[/tex]
[tex]x_{2} = \frac{-(-9) - \sqrt{9}}{2(1)} = 3[/tex]
6 and 3 with multiplicity 1, thus option B.
A similar problem is given at https://brainly.com/question/19776811
