Solve 40-8*x^2-8x-16/x^2-16

$$40 - 8 \cdot \frac { x ^ { 2 } - 8 x - 16 } { x ^ { 2 } - 16 }$$

Evaluate

$$32+\frac{64x}{x^{2}-16}$$

Short Solution Steps

Express 8\times \frac{x^{2}-8x-16}{x^{2}-16} as a single fraction.

$$40-\frac{8\left(x^{2}-8x-16\right)}{x^{2}-16}$$

Use the distributive property to multiply 8 by x^{2}-8x-16.

$$40-\frac{8x^{2}-64x-128}{x^{2}-16}$$

Factor x^{2}-16.

$$40-\frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 40 times \frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}.

$$\frac{40\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)}$$

Since \frac{40\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)} and \frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{40\left(x-4\right)\left(x+4\right)-\left(8x^{2}-64x-128\right)}{\left(x-4\right)\left(x+4\right)}$$

Do the multiplications in 40\left(x-4\right)\left(x+4\right)-\left(8x^{2}-64x-128\right).

$$\frac{40x^{2}+160x-160x-640-8x^{2}+64x+128}{\left(x-4\right)\left(x+4\right)}$$

Combine like terms in 40x^{2}+160x-160x-640-8x^{2}+64x+128.

$$\frac{32x^{2}+64x-512}{\left(x-4\right)\left(x+4\right)}$$

Expand \left(x-4\right)\left(x+4\right).

$$\frac{32x^{2}+64x-512}{x^{2}-16}$$

Expand

$$\frac{32\left(x^{2}+2x-16\right)}{x^{2}-16}$$

Short Solution Steps

Express 8\times \frac{x^{2}-8x-16}{x^{2}-16} as a single fraction.

$$40-\frac{8\left(x^{2}-8x-16\right)}{x^{2}-16}$$

Use the distributive property to multiply 8 by x^{2}-8x-16.

$$40-\frac{8x^{2}-64x-128}{x^{2}-16}$$

Factor x^{2}-16.

$$40-\frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 40 times \frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}.

$$\frac{40\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)}$$

Since \frac{40\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)} and \frac{8x^{2}-64x-128}{\left(x-4\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{40\left(x-4\right)\left(x+4\right)-\left(8x^{2}-64x-128\right)}{\left(x-4\right)\left(x+4\right)}$$

Do the multiplications in 40\left(x-4\right)\left(x+4\right)-\left(8x^{2}-64x-128\right).

$$\frac{40x^{2}+160x-160x-640-8x^{2}+64x+128}{\left(x-4\right)\left(x+4\right)}$$

Combine like terms in 40x^{2}+160x-160x-640-8x^{2}+64x+128.

$$\frac{32x^{2}+64x-512}{\left(x-4\right)\left(x+4\right)}$$

Expand \left(x-4\right)\left(x+4\right).

$$\frac{32x^{2}+64x-512}{x^{2}-16}$$