Solve (y - 2)/(y ^ 2 - 2y + 4) + (y + 2)/(y ^ 2 + 2y + 4) - 16/(y ^ 4 + 4y ^ 2 + 16) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{y-2}{y^{2}-2y+4}+\frac{y+2}{y^{2}+2y+4}-\frac{16}{y^{4}+4y^{2}+16}$$

Evaluate

$\frac{2\left(y-2\right)}{y^{2}-2y+4}$

Short Solution Steps

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

$$\frac{\left(y-2\right)\left(y^{2}+2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}+\frac{\left(y+2\right)\left(y^{2}-2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

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

$$\frac{\left(y-2\right)\left(y^{2}+2y+4\right)+\left(y+2\right)\left(y^{2}-2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Do the multiplications in $\left(y-2\right)\left(y^{2}+2y+4\right)+\left(y+2\right)\left(y^{2}-2y+4\right)$.

$$\frac{y^{3}+2y^{2}+4y-2y^{2}-4y-8+y^{3}-2y^{2}+4y+2y^{2}-4y+8}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Combine like terms in $y^{3}+2y^{2}+4y-2y^{2}-4y-8+y^{3}-2y^{2}+4y+2y^{2}-4y+8$.

$$\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Factor $y^{4}+4y^{2}+16$.

$$\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Since $\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$ and $\frac{16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2y^{3}-16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Factor the expressions that are not already factored in $\frac{2y^{3}-16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$.

$$\frac{2\left(y-2\right)\left(y^{2}+2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Cancel out $y^{2}+2y+4$ in both numerator and denominator.

$$\frac{2\left(y-2\right)}{y^{2}-2y+4}$$

Use the distributive property to multiply $2$ by $y-2$.

$$\frac{2y-4}{y^{2}-2y+4}$$

Expand

$\frac{2\left(y-2\right)}{y^{2}-2y+4}$

Short Solution Steps

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

$$\frac{\left(y-2\right)\left(y^{2}+2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}+\frac{\left(y+2\right)\left(y^{2}-2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

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

$$\frac{\left(y-2\right)\left(y^{2}+2y+4\right)+\left(y+2\right)\left(y^{2}-2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Do the multiplications in $\left(y-2\right)\left(y^{2}+2y+4\right)+\left(y+2\right)\left(y^{2}-2y+4\right)$.

$$\frac{y^{3}+2y^{2}+4y-2y^{2}-4y-8+y^{3}-2y^{2}+4y+2y^{2}-4y+8}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Combine like terms in $y^{3}+2y^{2}+4y-2y^{2}-4y-8+y^{3}-2y^{2}+4y+2y^{2}-4y+8$.

$$\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{y^{4}+4y^{2}+16}$$

Factor $y^{4}+4y^{2}+16$.

$$\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}-\frac{16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Since $\frac{2y^{3}}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$ and $\frac{16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2y^{3}-16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Factor the expressions that are not already factored in $\frac{2y^{3}-16}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$.

$$\frac{2\left(y-2\right)\left(y^{2}+2y+4\right)}{\left(y^{2}-2y+4\right)\left(y^{2}+2y+4\right)}$$

Cancel out $y^{2}+2y+4$ in both numerator and denominator.

$$\frac{2\left(y-2\right)}{y^{2}-2y+4}$$

Use the distributive property to multiply $2$ by $y-2$.

$$\frac{2y-4}{y^{2}-2y+4}$$