Answer:
(x^4) - 64x
Step-by-step explanation:
zeros at 4 and 0 are the roots so can be factor as x*(x-4) and the multiplicity tels to what power you have to raise those factors
[tex]x^{1} *(x-4)^{3}[/tex] becuase multiplicity of 1 for root at 0, and multiplicity of 3 at root 4
now to write this in standard form
[tex]x*(x-4)^{3} = \\x*(x-4)(x^{2} +4x+16)= \\(x^{2} -4x)(x^{2} +4x+16) = \\x^{4} +4x^{3} +16x^{2} -4x^{3} -16x^{2} -64x = \\x^{4} -64x[/tex]
