which of the following describes the zeroes of the graph of $f(x) = -x^5 + 9x^4 - 18x^3$?\n○ 0 with…

Explanation:

Step1: Factor the polynomial

First, factor out the greatest common factor (GCF) from the polynomial ( f(x) = -x^5 + 9x^4 - 18x^3 ). The GCF of the terms is ( -x^3 ), so we have: [ f(x) = -x^3(x^2 - 9x + 18) ]

Step2: Factor the quadratic

Next, factor the quadratic ( x^2 - 9x + 18 ). We need two numbers that multiply to 18 and add up to -9. Those numbers are -3 and -6? Wait, no, wait: actually, we need two numbers that multiply to 18 and add up to -9? Wait, no, the quadratic is ( x^2 - 9x + 18 ), so we need two numbers that multiply to 18 and add up to -9? Wait, no, the middle term is -9x, so the numbers should be -3 and -6? Wait, no, (-3)*(-6)=18 and (-3)+(-6)=-9. Wait, but let's check: ( (x - 3)(x - 6) = x^2 - 6x - 3x + 18 = x^2 - 9x + 18 ). Yes, that's correct. So the quadratic factors as ( (x - 3)(x - 6) ).

Step3: Write the fully factored form

Now, substitute back into the factored form of ( f(x) ): [ f(x) = -x^3(x - 3)(x - 6) ]

Step4: Identify the zeros and their multiplicities

To find the zeros, we set ( f(x) = 0 ): [ -x^3(x - 3)(x - 6) = 0 ] This gives us the solutions ( x^3 = 0 ), ( x - 3 = 0 ), and ( x - 6 = 0 ). Solving these, we get ( x = 0 ) (with multiplicity 3, since the exponent of ( x ) in ( -x^3 ) is 3), ( x = 3 ) (with multiplicity 1, since the exponent of ( (x - 3) ) is 1), and ( x = 6 ) (with multiplicity 1, since the exponent of ( (x - 6) ) is 1).

Answer:

0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1 (the second option)