Let $f (x) = 6 x^{2} + 7 x - 5$ $f(x) = 6x^2 + 7x - 5$ and g(x) = 2x - 1, how do you find f/g?

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Answer 1 · 2017-05-26T20:17:33

See a solution process below:

$(\frac{f}{g}) (x) = \frac{6 x^{2} + 7 x - 6}{2 x - 1}$ $(f/g)(x) = (6x^2 + 7x - 6)/(2x - 1)$

We can then factor the numerator:

$(\frac{f}{g}) (x) = \frac{(2 x - 1) (3 x + 5)}{2 x - 1}$ $(f/g)(x) = ((2x - 1)(3x + 5))/(2x - 1)$

We can now cancel common terms in the numerator and denominator:

$(f/g)(x) = (color(red)(cancel(color(black)((2x - 1))))(3x + 5))/color(red)(cancel(color(black)(2x - 1)))$

$(f/g)(x) = 3x + 5$

Where:

$(2x - 1) != 0$

Or

$x != 1/2$

Let f(x) = 6x^2 + 7x - 5f(x)=6x2+7x−5f(x) = 6x^2 + 7x - 5 and g(x) = 2x - 1, how do you find f/g?