Solve x^2-2x-9=A(x^2+3)+(Bx+C)(2x-1)

$$x ^ { 2 } - 2 x - 9 = A ( x ^ { 2 } + 3 ) + ( B x + C ) ( 2 x - 1 )$$

Solve for A

$$A=-\frac{2Bx^{2}-x^{2}-Bx+2Cx+2x-C+9}{x^{2}+3}$$

Steps for Solving Linear Equation

Use the distributive property to multiply A by x^{2}+3.

$$x^{2}-2x-9=Ax^{2}+3A+\left(Bx+C\right)\left(2x-1\right)$$

Use the distributive property to multiply Bx+C by 2x-1.

$$x^{2}-2x-9=Ax^{2}+3A+2Bx^{2}-Bx+2Cx-C$$

Swap sides so that all variable terms are on the left hand side.

$$Ax^{2}+3A+2Bx^{2}-Bx+2Cx-C=x^{2}-2x-9$$

Subtract 2Bx^{2} from both sides.

$$Ax^{2}+3A-Bx+2Cx-C=x^{2}-2x-9-2Bx^{2}$$

Add Bx to both sides.

$$Ax^{2}+3A+2Cx-C=x^{2}-2x-9-2Bx^{2}+Bx$$

Subtract 2Cx from both sides.

$$Ax^{2}+3A-C=x^{2}-2x-9-2Bx^{2}+Bx-2Cx$$

Add C to both sides.

$$Ax^{2}+3A=x^{2}-2x-9-2Bx^{2}+Bx-2Cx+C$$

Combine all terms containing A.

$$\left(x^{2}+3\right)A=x^{2}-2x-9-2Bx^{2}+Bx-2Cx+C$$

The equation is in standard form.

$$\left(x^{2}+3\right)A=-2Bx^{2}+x^{2}+Bx-2Cx-2x+C-9$$

Divide both sides by x^{2}+3.

$$\frac{\left(x^{2}+3\right)A}{x^{2}+3}=\frac{-2Bx^{2}+x^{2}+Bx-2Cx-2x+C-9}{x^{2}+3}$$

Dividing by x^{2}+3 undoes the multiplication by x^{2}+3.

$$A=\frac{-2Bx^{2}+x^{2}+Bx-2Cx-2x+C-9}{x^{2}+3}$$

Solve for B

$$\left\{\begin{matrix}B=-\frac{Ax^{2}-x^{2}+2Cx+2x+3A-C+9}{x\left(2x-1\right)}\text{, }&x\neq \frac{1}{2}\text{ and }x\neq 0\\B\in \mathrm{R}\text{, }&\left(A=\frac{C}{3}-3\text{ and }x=0\right)\text{ or }\left(A=-3\text{ and }x=\frac{1}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply A by x^{2}+3.

$$x^{2}-2x-9=Ax^{2}+3A+\left(Bx+C\right)\left(2x-1\right)$$

Use the distributive property to multiply Bx+C by 2x-1.

$$x^{2}-2x-9=Ax^{2}+3A+2Bx^{2}-Bx+2Cx-C$$

Swap sides so that all variable terms are on the left hand side.

$$Ax^{2}+3A+2Bx^{2}-Bx+2Cx-C=x^{2}-2x-9$$

Subtract Ax^{2} from both sides.

$$3A+2Bx^{2}-Bx+2Cx-C=x^{2}-2x-9-Ax^{2}$$

Subtract 3A from both sides.

$$2Bx^{2}-Bx+2Cx-C=x^{2}-2x-9-Ax^{2}-3A$$

Subtract 2Cx from both sides.

$$2Bx^{2}-Bx-C=x^{2}-2x-9-Ax^{2}-3A-2Cx$$

Add C to both sides.

$$2Bx^{2}-Bx=x^{2}-2x-9-Ax^{2}-3A-2Cx+C$$

Reorder the terms.

$$2Bx^{2}-Bx=-Ax^{2}+x^{2}-2Cx-2x+C-3A-9$$

Combine all terms containing B.

$$\left(2x^{2}-x\right)B=-Ax^{2}+x^{2}-2Cx-2x+C-3A-9$$

Divide both sides by 2x^{2}-x.

$$\frac{\left(2x^{2}-x\right)B}{2x^{2}-x}=\frac{-Ax^{2}+x^{2}-2Cx-2x+C-3A-9}{2x^{2}-x}$$

Dividing by 2x^{2}-x undoes the multiplication by 2x^{2}-x.

$$B=\frac{-Ax^{2}+x^{2}-2Cx-2x+C-3A-9}{2x^{2}-x}$$

Divide -Ax^{2}+x^{2}-2Cx-2x+C-3A-9 by 2x^{2}-x.

$$B=\frac{-Ax^{2}+x^{2}-2Cx-2x+C-3A-9}{x\left(2x-1\right)}$$