Solve 20/x^2+4+4/1

$$\frac{ 20 }{ { x }^{ 2 } +4 } + \frac{ 4 }{ 1 }$$

Evaluate

$$\frac{4\left(x^{2}+9\right)}{x^{2}+4}$$

Short Solution Steps

Anything divided by one gives itself.

$$\frac{20}{x^{2}+4}+4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{x^{2}+4}{x^{2}+4}.

$$\frac{20}{x^{2}+4}+\frac{4\left(x^{2}+4\right)}{x^{2}+4}$$

Since \frac{20}{x^{2}+4} and \frac{4\left(x^{2}+4\right)}{x^{2}+4} have the same denominator, add them by adding their numerators.

$$\frac{20+4\left(x^{2}+4\right)}{x^{2}+4}$$

Do the multiplications in 20+4\left(x^{2}+4\right).

$$\frac{20+4x^{2}+16}{x^{2}+4}$$

Combine like terms in 20+4x^{2}+16.

$$\frac{36+4x^{2}}{x^{2}+4}$$

Factor

$$\frac{4\left(x^{2}+9\right)}{x^{2}+4}$$