Question

Find the zeros for the given polynomial function and give the multiplicity for each zero. State whether the graph crosses the​ x-axis or touches the ​x-axis and turns around at each zero.
​f(x)equals x cubed minus 22 x squared plus 121 x
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Part 1
Determine the​ zero(s), if they exist.
The​ zero(s) is/are
  
enter your response here.
​(Type integers or decimals. Use a comma to separate answers as​ needed.)

Answer 1

To find the zeros of the polynomial function \( f(x) = x^3 - 22x^2 + 121x \), we first need to set \( f(x) = 0 \):

\[
x^3 - 22x^2 + 121x = 0
\]

Next, we can factor out \( x \):

\[
x(x^2 - 22x + 121) = 0
\]

Now, we have two factors: \( x = 0 \) and the quadratic \( x^2 - 22x + 121 \).

Next, we will find the roots of the quadratic \( x^2 - 22x + 121 \). We can use the quadratic formula where \( a = 1 \), \( b = -22 \), and \( c = 121 \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Calculating the discriminant:

\[
b^2 - 4ac = (-22)^2 - 4(1)(121) = 484 - 484 = 0
\]

Since the discriminant is zero, there is one repeated root for the quadratic:

\[
x = \frac{22}{2} = 11
\]

Thus, the zeros of the original polynomial \( f(x) \) are:

- \( x = 0 \) (from the factor \( x \))
- \( x = 11 \) (with a multiplicity of 2 from the quadratic \( (x - 11)^2 \))

To summarize, the zeros of \( f(x) \) are:

- \( 0 \) (multiplicity 1)
- \( 11 \) (multiplicity 2)

Now, let's analyze whether the graph crosses or touches the x-axis at each zero:

- At \( x = 0 \) (multiplicity 1): the graph **crosses** the x-axis.
- At \( x = 11 \) (multiplicity 2): the graph **touches** the x-axis and then turns around.

### Final Answer:
The zeros are: \( 0, 11 \)
The multiplicity for each zero:
- \( 0 \): multiplicity 1 (crosses the x-axis)
- \( 11 \): multiplicity 2 (touches the x-axis and turns around)