When #f(n) = n^2+3n# and #g(n) = n-1#, what is #(f * g)(n)#?

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Answer 1 · 2017-08-16T22:07:34

See a solution process below:

We can write #(f * g)(n)# as:

#(f * g)(n) = (n^2 + 3n) * (n - 1)#

We can next multiply each term in the parenthesis by each term in the parenthesis on the right giving:

#(f * g)(n) = (n^2 * n) - (n^2 * 1) + (3n * n) - (3n * 1)#

#(f * g)(n) = n^3 - n^2 + 3n^2 - 3n#

Now, we can combine like terms:

#(f * g)(n) = n^3 - 1n^2 + 3n^2 - 3n#

#(f * g)(n) = n^3 + (-1 + 3)n^2 - 3n#

#(f * g)(n) = n^3 + 2n^2 - 3n#