If $f (x) = x^{2} - 81$ $f(x)=x^2-81$ and $g (x) = {(x - 9)}^{- 1} (x + 9)$ $g(x)=(x-9)^-1(x+9)$ , how do you find $g (x) \times f (x)$ $g(x)timesf(x)$ ?

Question

5862 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-12T22:02:40

$g (x) \times f (x) = {(x + 9)}^{2}$ $g(x) xx f(x) = ( x + 9)^2$

$g (x) \times f (x) = x^{2} - 81 \times \frac{x + 9}{x - 9}$ $g(x) xx f(x) = x^2 - 81 xx (x + 9)/(x - 9)$

$= \frac{(x + 9) (x - 9) (x + 9)}{x - 9}$ $= ((x + 9)(x - 9)(x+ 9))/(x- 9)$

$= (x + 9) (x + 9)$ $=(x + 9)(x + 9)$

$= {(x + 9)}^{2}$ $=(x + 9)^2$

Hopefully this helps!

If f(x)=x2−81f(x)=x^2-81 and g(x)=(x−9)−1(x+9)g(x)=(x-9)^-1(x+9), how do you find g(x)×f(x)g(x)timesf(x)?