Write in Standard Form x^2-2x-y-8=0

$x^{2} - 2 x - y - 8 = 0$

Step 1

Solve for $y$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$- 2 x - y - 8 = - x^{2}$

Add $2 x$ to both sides of the equation.

$- y - 8 = - x^{2} + 2 x$

Add $8$ to both sides of the equation.

$- y = - x^{2} + 2 x + 8$

Divide each term in $- y = - x^{2} + 2 x + 8$ by $- 1$ and simplify.

Divide each term in $- y = - x^{2} + 2 x + 8$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- x^{2}}{- 1} + \frac{2 x}{- 1} + \frac{8}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- x^{2}}{- 1} + \frac{2 x}{- 1} + \frac{8}{- 1}$

Divide $y$ by $1$ .

$y = \frac{- x^{2}}{- 1} + \frac{2 x}{- 1} + \frac{8}{- 1}$

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

$y = \frac{x^{2}}{1} + \frac{2 x}{- 1} + \frac{8}{- 1}$

Divide $x^{2}$ by $1$ .

$y = x^{2} + \frac{2 x}{- 1} + \frac{8}{- 1}$

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

$y = x^{2} - 1 \cdot (2 x) + \frac{8}{- 1}$

Rewrite $- 1 \cdot (2 x)$ as $- (2 x)$ .

$y = x^{2} - (2 x) + \frac{8}{- 1}$

Multiply $2$ by $- 1$ .

$y = x^{2} - 2 x + \frac{8}{- 1}$

Divide $8$ by $- 1$ .

$y = x^{2} - 2 x - 8$

Step 2

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

Step 3

The standard form is $x^{2} - 2 x - 8$ .

$y = x^{2} - 2 x - 8$

Step 4