Question

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

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

Answer 1

See a solution process below:

Explanation:

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`