Solve (8x^4+6x^2-3x+1)div2x^2-x+2

$$(8 { x }^{ 4 } +6 { x }^{ 2 } -3x+1) \div 2 { x }^{ 2 } -x+2$$

Evaluate

$$4x^{6}+3x^{4}-\frac{3x^{3}}{2}+\frac{x^{2}}{2}-x+2$$

Short Solution Steps

Express \frac{8x^{4}+6x^{2}-3x+1}{2}x^{2} as a single fraction.

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2}-x+2$$

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

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2}+\frac{2\left(-x+2\right)}{2}$$

Since \frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2} and \frac{2\left(-x+2\right)}{2} have the same denominator, add them by adding their numerators.

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}+2\left(-x+2\right)}{2}$$

Do the multiplications in \left(8x^{4}+6x^{2}-3x+1\right)x^{2}+2\left(-x+2\right).

$$\frac{8x^{6}+6x^{4}-3x^{3}+x^{2}-2x+4}{2}$$

Expand

$$4x^{6}+3x^{4}-\frac{3x^{3}}{2}+\frac{x^{2}}{2}-x+2$$

Short Solution Steps

Express \frac{8x^{4}+6x^{2}-3x+1}{2}x^{2} as a single fraction.

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2}-x+2$$

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

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2}+\frac{2\left(-x+2\right)}{2}$$

Since \frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}}{2} and \frac{2\left(-x+2\right)}{2} have the same denominator, add them by adding their numerators.

$$\frac{\left(8x^{4}+6x^{2}-3x+1\right)x^{2}+2\left(-x+2\right)}{2}$$

Do the multiplications in \left(8x^{4}+6x^{2}-3x+1\right)x^{2}+2\left(-x+2\right).

$$\frac{8x^{6}+6x^{4}-3x^{3}+x^{2}-2x+4}{2}$$