Solve f(x)=ax^3-bx^2+23x-b=(x-2)(x-3)-g(x)

$$f ( x ) = a x ^ { 3 } - b x ^ { 2 } + 23 x - b = ( x - 2 ) ( x - 3 ) - g ( x )$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{-bx^{2}-x^{2}+gx+28x-b-6}{x^{3}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&b=-6\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-2 by x-3 and combine like terms.

$$ax^{3}-bx^{2}+23x-b=x^{2}-5x+6-gx$$

Subtract 23x from both sides.

$$ax^{3}-bx^{2}-b=x^{2}-5x+6-gx-23x$$

Combine -5x and -23x to get -28x.

$$ax^{3}-bx^{2}-b=x^{2}-28x+6-gx$$

Add b to both sides.

$$ax^{3}-bx^{2}=x^{2}-28x+6-gx+b$$

Add bx^{2} to both sides.

$$ax^{3}=x^{2}-28x+6-gx+b+bx^{2}$$

The equation is in standard form.

$$x^{3}a=bx^{2}+x^{2}-gx-28x+b+6$$

Divide both sides by x^{3}.

$$\frac{x^{3}a}{x^{3}}=\frac{bx^{2}+x^{2}-gx-28x+b+6}{x^{3}}$$

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

$$a=\frac{bx^{2}+x^{2}-gx-28x+b+6}{x^{3}}$$

Solve for b

$$b=\frac{ax^{3}-x^{2}+gx+28x-6}{x^{2}+1}$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-2 by x-3 and combine like terms.

$$ax^{3}-bx^{2}+23x-b=x^{2}-5x+6-gx$$

Subtract 23x from both sides.

$$ax^{3}-bx^{2}-b=x^{2}-5x+6-gx-23x$$

Combine -5x and -23x to get -28x.

$$ax^{3}-bx^{2}-b=x^{2}-28x+6-gx$$

Subtract ax^{3} from both sides.

$$-bx^{2}-b=x^{2}-28x+6-gx-ax^{3}$$

Combine all terms containing b.

$$\left(-x^{2}-1\right)b=x^{2}-28x+6-gx-ax^{3}$$

The equation is in standard form.

$$\left(-x^{2}-1\right)b=6-28x-gx+x^{2}-ax^{3}$$

Divide both sides by -x^{2}-1.

$$\frac{\left(-x^{2}-1\right)b}{-x^{2}-1}=\frac{6-28x-gx+x^{2}-ax^{3}}{-x^{2}-1}$$

Dividing by -x^{2}-1 undoes the multiplication by -x^{2}-1.

$$b=\frac{6-28x-gx+x^{2}-ax^{3}}{-x^{2}-1}$$

Divide x^{2}-28x+6-gx-ax^{3} by -x^{2}-1.

$$b=-\frac{6-28x-gx+x^{2}-ax^{3}}{x^{2}+1}$$