Given $f (x) = x + 2$ $f(x) = x +2$ and $g (x) = 2 x^{2} - 4 x + 2$ $g(x)= 2x^2-4x+2$ , how do you find $f (x) \div g (x)$ $f(x) ÷ g(x)$ ?

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Answer 1 · 2017-03-24T16:59:17

$\frac{1}{2} \frac{x + 2}{{(x - 1)}^{2}}$ $1/2(x+2)/(x-1)^2$

$\frac{f (x)}{g (x)} = \frac{x + 2}{2 x^{2} - 4 x + 2}$ $f(x)/g(x) = [x+2]/[2x^2 - 4x +2]$

$\Rightarrow \frac{x + 2}{2 {x^{2} - 2 x + 1}}$ $rArr[x+2]/[2{x^2-2x+1}]$

$\Rightarrow \frac{1}{2} \frac{x + 2}{{(x - 1)}^{2}}$ $rArr 1/2(x+2)/(x-1)^2$

Given f(x)=x+2f(x) = x +2 and g(x)=2x2−4x+2g(x)= 2x^2-4x+2, how do you find f(x)÷g(x)f(x) ÷ g(x)?