Solve (4/2z+1+2/1+z-2z^2)*2z+3z+1/3+z-2z^2

$$( \frac { 4 } { 2 z + 1 } + \frac { 2 } { 1 + z - 2 z ^ { 2 } } ) \cdot \frac { 2 z + 3 z + 1 } { 3 + z - 2 z ^ { 2 } }$$

Evaluate

$$-\frac{2\left(5z+1\right)}{\left(2z+1\right)\left(z^{2}-1\right)}$$

Short Solution Steps

Factor 1+z-2z^{2}.

$$\left(\frac{4}{2z+1}+\frac{2}{\left(-z+1\right)\left(2z+1\right)}\right)\times \frac{2z+3z+1}{3+z-2z^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2z+1 and \left(-z+1\right)\left(2z+1\right) is \left(-z+1\right)\left(2z+1\right). Multiply \frac{4}{2z+1} times \frac{-z+1}{-z+1}.

$$\left(\frac{4\left(-z+1\right)}{\left(-z+1\right)\left(2z+1\right)}+\frac{2}{\left(-z+1\right)\left(2z+1\right)}\right)\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Since \frac{4\left(-z+1\right)}{\left(-z+1\right)\left(2z+1\right)} and \frac{2}{\left(-z+1\right)\left(2z+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{4\left(-z+1\right)+2}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Do the multiplications in 4\left(-z+1\right)+2.

$$\frac{-4z+4+2}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Combine like terms in -4z+4+2.

$$\frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Combine 2z and 3z to get 5z.

$$\frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)}\times \frac{5z+1}{3+z-2z^{2}}$$

Multiply \frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)} times \frac{5z+1}{3+z-2z^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(-4z+6\right)\left(5z+1\right)}{\left(-z+1\right)\left(2z+1\right)\left(3+z-2z^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\left(-2z+3\right)\left(5z+1\right)}{\left(-z-1\right)\left(2z-3\right)\left(-z+1\right)\left(2z+1\right)}$$

Extract the negative sign in 3-2z.

$$\frac{-2\left(2z-3\right)\left(5z+1\right)}{\left(-z-1\right)\left(2z-3\right)\left(-z+1\right)\left(2z+1\right)}$$

Cancel out 2z-3 in both numerator and denominator.

$$\frac{-2\left(5z+1\right)}{\left(-z-1\right)\left(-z+1\right)\left(2z+1\right)}$$

Expand the expression.

$$\frac{-10z-2}{2z^{3}+z^{2}-2z-1}$$

Expand

$$-\frac{2\left(5z+1\right)}{\left(2z+1\right)\left(z^{2}-1\right)}$$

Short Solution Steps

Factor 1+z-2z^{2}.

$$\left(\frac{4}{2z+1}+\frac{2}{\left(-z+1\right)\left(2z+1\right)}\right)\times \frac{2z+3z+1}{3+z-2z^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2z+1 and \left(-z+1\right)\left(2z+1\right) is \left(-z+1\right)\left(2z+1\right). Multiply \frac{4}{2z+1} times \frac{-z+1}{-z+1}.

$$\left(\frac{4\left(-z+1\right)}{\left(-z+1\right)\left(2z+1\right)}+\frac{2}{\left(-z+1\right)\left(2z+1\right)}\right)\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Since \frac{4\left(-z+1\right)}{\left(-z+1\right)\left(2z+1\right)} and \frac{2}{\left(-z+1\right)\left(2z+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{4\left(-z+1\right)+2}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Do the multiplications in 4\left(-z+1\right)+2.

$$\frac{-4z+4+2}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Combine like terms in -4z+4+2.

$$\frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)}\times \frac{2z+3z+1}{3+z-2z^{2}}$$

Combine 2z and 3z to get 5z.

$$\frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)}\times \frac{5z+1}{3+z-2z^{2}}$$

Multiply \frac{-4z+6}{\left(-z+1\right)\left(2z+1\right)} times \frac{5z+1}{3+z-2z^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(-4z+6\right)\left(5z+1\right)}{\left(-z+1\right)\left(2z+1\right)\left(3+z-2z^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\left(-2z+3\right)\left(5z+1\right)}{\left(-z-1\right)\left(2z-3\right)\left(-z+1\right)\left(2z+1\right)}$$

Extract the negative sign in 3-2z.

$$\frac{-2\left(2z-3\right)\left(5z+1\right)}{\left(-z-1\right)\left(2z-3\right)\left(-z+1\right)\left(2z+1\right)}$$

Cancel out 2z-3 in both numerator and denominator.

$$\frac{-2\left(5z+1\right)}{\left(-z-1\right)\left(-z+1\right)\left(2z+1\right)}$$

Expand the expression.

$$\frac{-10z-2}{2z^{3}+z^{2}-2z-1}$$