Solve for C C(x)=-4x^2+4x+3

$C (x) = - 4 x^{2} + 4 x + 3$

Step 1

Divide each term in $C x = - 4 x^{2} + 4 x + 3$ by $x$ .

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

Step 2

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $C$ by $1$ .

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

Step 3

Simplify the right side.

Simplify each term.

Cancel the common factor of $x^{2}$ and $x$ .

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

$C = \frac{x (- 4 x)}{x} + \frac{4 x}{x} + \frac{3}{x}$

Cancel the common factors.

Raise $x$ to the power of $1$ .

$C = \frac{x (- 4 x)}{x^{1}} + \frac{4 x}{x} + \frac{3}{x}$

Factor $x$ out of $x^{1}$ .

$C = \frac{x (- 4 x)}{x \cdot 1} + \frac{4 x}{x} + \frac{3}{x}$

Cancel the common factor.

$C = \frac{x (- 4 x)}{x \cdot 1} + \frac{4 x}{x} + \frac{3}{x}$

Rewrite the expression.

$C = \frac{- 4 x}{1} + \frac{4 x}{x} + \frac{3}{x}$

Divide $- 4 x$ by $1$ .

$C = - 4 x + \frac{4 x}{x} + \frac{3}{x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$C = - 4 x + \frac{4 x}{x} + \frac{3}{x}$

Divide $4$ by $1$ .

$C = - 4 x + 4 + \frac{3}{x}$