Question

which of the following describes the zeroes of the graph of $f(x) = -x^5 + 9x^4 - 18x^3$?
\bigcirc 0 with multiplicity 3, $-3$ with multiplicity 2, and $-2$ with multiplicity 1
\bigcirc 0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1
\bigcirc 0 with multiplicity 3, 3 with multiplicity 2, and 2 with multiplicity 1
\bigcirc 0 with multiplicity 3, $-3$ with multiplicity 1, and $-6$ with multiplicity 1

Explanation:

Step1: Factor the polynomial

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

Step2: Factor the quadratic

Now, factor the quadratic \( x^2 - 9x + 18 \). We need two numbers that multiply to 18 and add to -9. Those numbers are -3 and -6? Wait, no, wait: actually, let's check again. Wait, the quadratic is \( x^2 - 9x + 18 \). Let's factor it:
We need two numbers \( a \) and \( b \) such that \( a + b = -9 \) and \( a \times b = 18 \). Wait, no, the quadratic is \( x^2 - 9x + 18 \), so the coefficient of \( x \) is -9, and the constant term is 18. So we need two numbers that multiply to 18 and add to -9. Wait, actually, 3 and 6: 3 + 6 = 9, so -3 and -6: (-3) + (-6) = -9, and (-3)×(-6) = 18. Wait, but then the quadratic would factor as \( (x - 3)(x - 6) \)? Wait, no: \( (x - 3)(x - 6) = x^2 - 9x + 18 \), yes! Wait, but let's check:
\( (x - 3)(x - 6) = x^2 - 6x - 3x + 18 = x^2 - 9x + 18 \). Yes, that's correct. Wait, but hold on, the original polynomial after factoring out \( -x^3 \) is \( -x^3(x^2 - 9x + 18) \), so factoring the quadratic gives \( -x^3(x - 3)(x - 6) \). Wait, but that would mean the roots are \( x = 0 \) (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). Wait, but let's check the options. The second option is "0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1". Let's verify again.
Wait, maybe I made a mistake in factoring. Let's re-express the original polynomial:
\( f(x) = -x^5 + 9x^4 - 18x^3 \)
Factor out \( -x^3 \):
\( f(x) = -x^3(x^2 - 9x + 18) \)
Now, factor \( x^2 - 9x + 18 \). Let's find two numbers that multiply to 18 and add to -9. Wait, 3 and 6: 3×6=18, 3+6=9. So to get -9x, we need -3 and -6: (-3) + (-6) = -9, (-3)×(-6)=18. So \( x^2 - 9x + 18 = (x - 3)(x - 6) \). Therefore, the polynomial factors as \( -x^3(x - 3)(x - 6) \). Therefore, the roots are \( x = 0 \) (multiplicity 3), \( x = 3 \) (multiplicity 1), and \( x = 6 \) (multiplicity 1). So the second option is correct.

Answer:

0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1