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:
x =0, x =4, x = 4
Step-by-step explanation:
Given: f(x) = \(x^3 -8x^2 +16x\)
To find the zeros, we need to plug in f(x) = 0, we get
\(x^3 - 8x^2 +16x = 0\)
Here x is the common factor, so we can take it out.
\(x(x^2 - 8x +16) = 0\)
Now we can factorize \(x^2 -8x +16\)
\(x^2 -8x +16 = (x -4)(x-4)\)
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