Answer:
0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1.
Factoring and solving a quadratic equation, it is found that the zeroes of the graph are:
0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1 .
The function given is:
[tex]f(x) = -x^5 + 9x^4 - 18x^3[/tex]
The zeroes are the values of x for which:
[tex]f(x) = 0[/tex]
[tex]-x^5 + 9x^4 - 18x^3 = 0[/tex]
Factoring by the common term:
[tex]-x^5 + 9x^4 - 18x^3 = -x^3(x^2 - 9x + 18) = 0[/tex]
Thus:
[tex]-x^3 = 0 \rightarrow x = 0[/tex]
0 has a multiplicity of 3.
[tex]x^2 - 9x + 18 = 0[/tex]
Which has coefficients [tex]a = 1, b = -9, c = 18[/tex], thus:
[tex]\Delta = (-9)^{2} - 4(1)(18) = 9[/tex]
[tex]x_{1} = \frac{-(-9) + \sqrt{9}}{2} = 6[/tex]
[tex]x_{2} = \frac{-(-9) - \sqrt{9}}{2} = 3[/tex]
6 and 3 with multiplicity 1.
Thus, the correct option is:
0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1 .
A similar problem is given at https://brainly.com/question/24380382
