What is the answer to solvefor g,g(r)=(r^2+17)(r^3-3r+1) ?

The answer to solvefor g,g(r)=(r^2+17)(r^3-3r+1) is g=((r^2+17)(r^3-3r+1))/r ;\quad r\ne 0

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solvefor g,g(r)=(r^2+17)(r^3-3r+1)

solve for

g, g (r) = (r^{2} + 17) (r^{3} - 3 r + 1)

g = \frac{( r ^{2} + 17 ) ( r ^{3} - 3 r + 1 )}{r}; r \neq = 0

Solution steps

g (r) = (r^{2} + 17) (r^{3} - 3 r + 1)

Divide both sides by

r; r \neq = 0

g = \frac{( r ^{2} + 17 ) ( r ^{3} - 3 r + 1 )}{r}; r \neq = 0

\frac{4}{2} = \frac{x ( 1 + 4 )}{5}

x, f (f (x - 1)) = 16 x - 1

7 + 5 x - 2 = x - 3 + 2 x

6 = \frac{2}{3} \cdot (- 2) + c

5 y = 4

What is the answer to solvefor g,g(r)=(r^2+17)(r^3-3r+1) ?
The answer to solvefor g,g(r)=(r^2+17)(r^3-3r+1) is g=((r^2+17)(r^3-3r+1))/r ;\quad r\ne 0