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 $x != - 4$

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