Solve =x^3+19x^2+98x+80quadyquadq(x)=x^2+9x+8

$$= x ^ { 3 } + 19 x ^ { 2 } + 98 x + 80 \quad y \quad q ( x ) = x ^ { 2 } + 9 x + 8$$

Solve for q

$$\left\{\begin{matrix}q=-\frac{x^{3}+18x^{2}+89x-8}{80xy}\text{, }&x\neq 0\text{ and }y\neq 0\\q\in \mathrm{R}\text{, }&x=\sqrt[3]{\frac{8\sqrt{3507}}{9}+55}+\sqrt[3]{-\frac{8\sqrt{3507}}{9}+55}-6\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract x^{3} from both sides.

$$19x^{2}+98x+80yqx=x^{2}+9x+8-x^{3}$$

Subtract 19x^{2} from both sides.

$$98x+80yqx=x^{2}+9x+8-x^{3}-19x^{2}$$

Combine x^{2} and -19x^{2} to get -18x^{2}.

$$98x+80yqx=-18x^{2}+9x+8-x^{3}$$

Subtract 98x from both sides.

$$80yqx=-18x^{2}+9x+8-x^{3}-98x$$

Combine 9x and -98x to get -89x.

$$80yqx=-18x^{2}-89x+8-x^{3}$$

The equation is in standard form.

$$80xyq=8-89x-18x^{2}-x^{3}$$

Divide both sides by 80yx.

$$\frac{80xyq}{80xy}=\frac{8-89x-18x^{2}-x^{3}}{80xy}$$

Dividing by 80yx undoes the multiplication by 80yx.

$$q=\frac{8-89x-18x^{2}-x^{3}}{80xy}$$