Question

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

Explanation:

Step1: Factor the function

To find the zeroes of \( f(x) = -x^5 + 9x^4 - 18x^3 \), we first factor out the greatest common factor, which is \( -x^3 \). So we have:
\( f(x) = -x^3(x^2 - 9x + 18) \)

Step2: Factor the quadratic

Next, we 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. So the quadratic factors as:
\( x^2 - 9x + 18 = (x - 3)(x - 6) \)

Step3: Write the fully factored form

Putting it all together, the fully factored form of the function is:
\( f(x) = -x^3(x - 3)(x - 6) \)

Step4: Identify the zeroes and their multiplicities

The zeroes of the function are the values of \( x \) that make \( f(x) = 0 \). From the factored form:

• \( -x^3 = 0 \) gives \( x = 0 \). The exponent of \( x \) here is 3, so the multiplicity of 0 is 3.
• \( (x - 3) = 0 \) gives \( x = 3 \). The exponent of \( (x - 3) \) is 1, so the multiplicity of 3 is 1.
• \( (x - 6) = 0 \) gives \( x = 6 \). The exponent of \( (x - 6) \) is 1, so the multiplicity of 6 is 1.

Answer:

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