What is the answer to solvefor f,(f*g)(x)=(x^2-3x+2)(x-1) ?

The answer to solvefor f,(f*g)(x)=(x^2-3x+2)(x-1) is f=((x^2-3x+2)(x-1))/(xg);\quad x\ne 0,g\ne 0

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solvefor f,(f*g)(x)=(x^2-3x+2)(x-1)

solve for

f, (f \cdot g) (x) = (x^{2} - 3 x + 2) (x - 1)

f = \frac{( x ^{2} - 3 x + 2 ) ( x - 1 )}{xg}; x \neq = 0, g \neq = 0

Solution steps

(f g) (x) = (x^{2} - 3 x + 2) (x - 1)

Divide both sides by

x; x \neq = 0

f g = \frac{( x ^{2} - 3 x + 2 ) ( x - 1 )}{x}; x \neq = 0

Divide both sides by

g; x \neq = 0, g \neq = 0

f = \frac{( x ^{2} - 3 x + 2 ) ( x - 1 )}{xg}; x \neq = 0, g \neq = 0

6 (x - 1) + 21 = 6 x + 15

x - 7 = 7

x + x = 0

\frac{3}{5} (x + 2) = \frac{3}{2} (2 x - 4) - \frac{2}{5}

x, 4 x + 4 y = 16

What is the answer to solvefor f,(f*g)(x)=(x^2-3x+2)(x-1) ?
The answer to solvefor f,(f*g)(x)=(x^2-3x+2)(x-1) is f=((x^2-3x+2)(x-1))/(xg);\quad x\ne 0,g\ne 0