Given f(x)=3x2+10x−8 and g(x)=3x2−2x .

What is (f/g)(x) ?

−5x+4/x where x≠0, 2/3

x/−5x+4 where x≠0, 4/5

x/x+4 where x≠0, 4

x+4/x where x≠0, 2/3

BellaNight7310 May 13, 2019

#2**+3 **

Given \(f(x)=3x^2+10x-8\) and \(g(x)=3x^2-2x\)

What is \((\frac{f}{g})(x)\) ?

--------------------------

\((\frac{f}{g})(x)\,=\,\frac{f(x)}{g(x)}\)

Substitute \(3x^2+10x-8\) in for \(f(x)\) and substitute \(3x^2-2x\) in for \(g(x)\)

\((\frac{f}{g})(x)\,=\,\frac{3x^2+10x-8}{3x^2-2x}\)

Split 10x into two terms such that their coefficients add to 10 and multiply to -24

\((\frac{f}{g})(x)\,=\,\frac{3x^2+12x-2x-8}{3x^2-2x}\)

Factor the numerator and denominator.

\((\frac{f}{g})(x)\,=\,\frac{3x(x+4)-2(x+4)}{3x^2-2x}\\~\\ (\frac{f}{g})(x)\,=\,\frac{(x+4)(3x-2)}{x(3x-2)} \)

Cancel the common factor of (3x -2) and note that 3x - 2 ≠ 0 , that is, x ≠ 2/3

\( (\frac{f}{g})(x)\,=\,\frac{x+4}{x}\qquad\text{and}\qquad x\neq\frac23\)

Since x is in the denominator we can also note that x ≠ 0

hectictar May 14, 2019