What is the multiple zero and multiplicity of f(x)=x^3-8x^2+16x?

multiplicity is how many times a root repeats
factor
x(x²-8x+16)
what times wat =16 and adds to -8
-4 and -4
x(x-4)(x-4)
x(x-4)²
set to zero
x=0

x-4=0
x=4

roots are 0 ad 4

4 repeats 2 times, so has a multiplicty of 2

so
roots are 0, and 4 multiplicity 2

Answer Link

ikarus ikarus · Answer 2 · 2019-06-24T20:26:32+02:00

Answer:

x =0, x =4, x = 4

Step-by-step explanation:

Given: f(x) = [tex]x^3 -8x^2 +16x[/tex]

To find the zeros, we need to plug in f(x) = 0, we get

[tex]x^3 - 8x^2 +16x = 0[/tex]

Here x is the common factor, so we can take it out.

[tex]x(x^2 - 8x +16) = 0[/tex]

Now we can factorize [tex]x^2 -8x +16[/tex]

[tex]x^2 -8x +16 = (x -4)(x-4)[/tex]

So

x(x^2 - 8x -16) = 0

x(x-4)(x-4) = 0

x = 0, (x -4) = 0, x-4 = 0

x =0, x = 4, x = 4

Here the two roots are real and equal.

Therefore, the zeros of the given function are x =0 and x = 4 and x = 4