Write in Standard Form 4x^2-128y=0

$4 x^{2} - 128 y = 0$

Step 1

Solve for $y$ .

Subtract $4 x^{2}$ from both sides of the equation.

$- 128 y = - 4 x^{2}$

Divide each term in $- 128 y = - 4 x^{2}$ by $- 128$ and simplify.

Divide each term in $- 128 y = - 4 x^{2}$ by $- 128$ .

$\frac{- 128 y}{- 128} = \frac{- 4 x^{2}}{- 128}$

Simplify the left side.

Cancel the common factor of $- 128$ .

Cancel the common factor.

$\frac{- 128 y}{- 128} = \frac{- 4 x^{2}}{- 128}$

Divide $y$ by $1$ .

$y = \frac{- 4 x^{2}}{- 128}$

Simplify the right side.

Cancel the common factor of $- 4$ and $- 128$ .

Factor $- 4$ out of $- 4 x^{2}$ .

$y = \frac{- 4 (x^{2})}{- 128}$

Cancel the common factors.

Factor $- 4$ out of $- 128$ .

$y = \frac{- 4 (x^{2})}{- 4 (32)}$

Cancel the common factor.

$y = \frac{- 4 x^{2}}{- 4 \cdot 32}$

Rewrite the expression.

$y = \frac{x^{2}}{32}$

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

Reorder terms.

$y = \frac{1}{32} x^{2}$

Step 4