Solve -8x^4-2x^2+xdiv4^2-1

$$-8 { x }^{ 4 } -2 { x }^{ 2 } +x \div { 4 }^{ 2 } -1$$

Evaluate

$$-8x^{4}-2x^{2}+\frac{x}{16}-1$$

Short Solution Steps

Calculate 4 to the power of 2 and get 16.

$$-8x^{4}-2x^{2}+\frac{x}{16}-1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -8x^{4}-2x^{2}-1 times \frac{16}{16}.

$$\frac{16\left(-8x^{4}-2x^{2}-1\right)}{16}+\frac{x}{16}$$

Since \frac{16\left(-8x^{4}-2x^{2}-1\right)}{16} and \frac{x}{16} have the same denominator, add them by adding their numerators.

$$\frac{16\left(-8x^{4}-2x^{2}-1\right)+x}{16}$$

Do the multiplications in 16\left(-8x^{4}-2x^{2}-1\right)+x.

$$\frac{-128x^{4}-32x^{2}-16+x}{16}$$

Factor

$$\frac{-128x^{4}-32x^{2}+x-16}{16}$$