Answer:
zeros
x=0
x=4 with multiplicity 2
Step-by-step explanation:
We need to solve x^3-8x^2+16x=0
Notice each term has a factor of x in common in x^3-8x^2+16x so we can factor it as x(x^2-8x+16)
Now x^2-8x+16 is a quadratic where a=1... We can see if it is factorable by looking for two numbers that multiply to be 16 and add up to be -8 which is -4 and -4
So you have x^3-8x^2+16x=0 is equivalent to x(x-4)(x-4)=0 (this one is in factored form).
x=0
x=4 (multiplicity 2 since you had the factor that is came from occurring twice)
