Answer. Define "f(n) = 1n^3+2n+3n^2". We try to factorize this polynomial. One obvious factor is "n", so "f(n) = n(1n^2+2+3n)". We factorize the factor of degree "2" by completing the square:
"= \\left(n +\\frac{3}{2} -\\frac{1}{2}\\right) \\left(n +\\frac{3}{2} +\\frac{1}{2}\\right) = (n+1)(n+2)."
Hence
"f(n) = n (n+1) (n+2)."Let "n" be a positive integer number. Division of "n" by "3" gives an integer quotient "q" and a remainder "r\\in \\{0, 1, 2\\}" such that "n = 3q+r". Now check all the possible values of "r".
Hence "3" divides "f(n)" in all cases.
Division of "n" by "2" gives an integer quotient "q" and a remainder "r\\in \\{0, 1\\}" such that "n = 2q+r".
Hence "2" divides "f(n)" in all cases.
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