What is the answer to (z+1)(z+2)=1+3z^{-1}+2z^{-2} ?

The answer to (z+1)(z+2)=1+3z^{-1}+2z^{-2} is z=-1,z=-2,z=1

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(z+1)(z+2)=1+3z^{-1}+2z^{-2}

(z + 1) (z + 2) = 1 + 3 z^{- 1} + 2 z^{- 2}

z = - 1, z = - 2, z = 1

Solution steps

(z + 1) (z + 2) = 1 + 3 z^{- 1} + 2 z^{- 2}

Refine

(z + 1) (z + 2) = 1 + \frac{3}{z} + \frac{2}{z ^{2}}

Multiply by LCM

(z + 1) (z + 2) z^{2} = z^{2} + 3 z + 2

Solve

(z + 1) (z + 2) z^{2} = z^{2} + 3 z + 2 : z = - 1, z = - 2, z = 1

z = - 1, z = - 2, z = 1

Verify Solutions

Find undefined (singularity) points:

z = 0

Combine undefined points with solutions:

z = - 1, z = - 2, z = 1

x, x^{- 3} + y^{3} = 27

\frac{2 ( - x ^{2} + 18 )}{x ^{3}} = - 5

- \frac{10}{p - 4} = \frac{3}{p - 6}

\frac{( 4 n - 2 )}{2 n} = 7

1 - \frac{16}{x ^{2}} = 0

What is the answer to (z+1)(z+2)=1+3z^{-1}+2z^{-2} ?
The answer to (z+1)(z+2)=1+3z^{-1}+2z^{-2} is z=-1,z=-2,z=1
Solve for z: (z+1)(z+2)=1+3z^{-1}+2z^{-2}
The solution is z=-1,z=-2,z=1