How do I find f #f*g# and #f/g# if #f(x)=(x^2+3x-4)# and #g(x)=(x+4)#?

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Answer 1 · 2018-01-04T13:10:13

See a solution process below:

First, we can factor #(x^2 +3x - 4)# as: #(x + 4)(x - 1)#

Therefore: #f * g# is:

#f * g = f(x) xx g(x) = (x^2 + 3x - 4)(x + 4) =#

#(x + 4)(x - 1)(x + 4) = (x + 4)^2(x - 1)#

Or, we can expand #(x^2 + 3x - 4)(x + 4)# as:

#f * g = (x^2 + 3x - 4)(x + 4)#

#f * g = (x^2 xx x) + (3x xx x) - (4 xx x) + (x^2 xx 4) + (3x xx 4) - (4 xx + 4)#

#f * g = x^3 + 3x^2 - 4x + 4x^2 + 12x - 16#

#f * g = x^3 + 3x^2 + 4x^2 - 4x+ 12x - 16#

#f * g = x^3 + 7x^2 + 8x - 16# or #f * g = (x + 4)^2(x - 1)#

Then, #f/g# is:

#f/g = f(x)/g(x) = (x^2 + 3x - 4)/(x + 4) =#

#((x + 4)(x - 1))/(x + 4) = (color(red)(cancel(color(black)((x + 4))))(x - 1))/color(red)(cancel(color(black)(x + 4)))#

#f/g = x - 1#

However, because we cannot divide by zero we must find the exclusions:

#x + 4 != 0# therefore #x != -4#

#f/g = x - 1# where #x != - 4#