Solve f(x)=(x-a)(x-b)(x-c),a,b,cinR,quadf^prime(x)forallf(x)

$$f ( x ) = ( x - a ) ( x - b ) ( x - c ) , a , b , c \in R , \quad f ^ { \prime } ( x ) \forall f ( x )$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{x\left(x^{2}-bx-cx+bc-f\right)}{\left(b-x\right)\left(x-c\right)}\text{, }&x\neq c\text{ and }x\neq b\\a\in \mathrm{R}\text{, }&\left(f=0\text{ and }x=c\right)\text{ or }\left(x=0\text{ and }c=0\right)\text{ or }\left(x=0\text{ and }b=0\text{ and }c\neq 0\right)\text{ or }\left(f=0\text{ and }x=b\text{ and }b\neq c\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-a by x-b.

$$fx=\left(x^{2}-xb-ax+ab\right)\left(x-c\right)$$

Use the distributive property to multiply x^{2}-xb-ax+ab by x-c.

$$fx=x^{3}-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc$$

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

$$x^{3}-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx$$

Subtract x^{3} from both sides.

$$-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx-x^{3}$$

Add x^{2}c to both sides.

$$-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx-x^{3}+x^{2}c$$

Add bx^{2} to both sides.

$$bxc-ax^{2}+axc+abx-abc=fx-x^{3}+x^{2}c+bx^{2}$$

Subtract bxc from both sides.

$$-ax^{2}+axc+abx-abc=fx-x^{3}+x^{2}c+bx^{2}-bxc$$

Reorder the terms.

$$-ax^{2}+abx+acx-abc=-x^{3}+bx^{2}+cx^{2}-bcx+fx$$

Combine all terms containing a.

$$\left(-x^{2}+bx+cx-bc\right)a=-x^{3}+bx^{2}+cx^{2}-bcx+fx$$

The equation is in standard form.

$$\left(-x^{2}+bx+cx-bc\right)a=fx-bcx+cx^{2}+bx^{2}-x^{3}$$

Divide both sides by bx-bc-x^{2}+xc.

$$\frac{\left(-x^{2}+bx+cx-bc\right)a}{-x^{2}+bx+cx-bc}=\frac{x\left(f-bc+cx+bx-x^{2}\right)}{-x^{2}+bx+cx-bc}$$

Dividing by bx-bc-x^{2}+xc undoes the multiplication by bx-bc-x^{2}+xc.

$$a=\frac{x\left(f-bc+cx+bx-x^{2}\right)}{-x^{2}+bx+cx-bc}$$

Divide x\left(-x^{2}+bx+cx-bc+f\right) by bx-bc-x^{2}+xc.

$$a=\frac{x\left(f-bc+cx+bx-x^{2}\right)}{\left(b-x\right)\left(x-c\right)}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{x\left(x^{2}-ax-cx+ac-f\right)}{\left(a-x\right)\left(x-c\right)}\text{, }&x\neq c\text{ and }x\neq a\\b\in \mathrm{R}\text{, }&\left(f=0\text{ and }x=c\right)\text{ or }\left(x=0\text{ and }c=0\right)\text{ or }\left(x=0\text{ and }a=0\text{ and }c\neq 0\right)\text{ or }\left(f=0\text{ and }x=a\text{ and }a\neq c\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-a by x-b.

$$fx=\left(x^{2}-xb-ax+ab\right)\left(x-c\right)$$

Use the distributive property to multiply x^{2}-xb-ax+ab by x-c.

$$fx=x^{3}-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc$$

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

$$x^{3}-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx$$

Subtract x^{3} from both sides.

$$-x^{2}c-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx-x^{3}$$

Add x^{2}c to both sides.

$$-bx^{2}+bxc-ax^{2}+axc+abx-abc=fx-x^{3}+x^{2}c$$

Add ax^{2} to both sides.

$$-bx^{2}+bxc+axc+abx-abc=fx-x^{3}+x^{2}c+ax^{2}$$

Subtract axc from both sides.

$$-bx^{2}+bxc+abx-abc=fx-x^{3}+x^{2}c+ax^{2}-axc$$

Reorder the terms.

$$-bx^{2}+abx+bcx-abc=-x^{3}+ax^{2}+cx^{2}-acx+fx$$

Combine all terms containing b.

$$\left(-x^{2}+ax+cx-ac\right)b=-x^{3}+ax^{2}+cx^{2}-acx+fx$$

The equation is in standard form.

$$\left(-x^{2}+ax+cx-ac\right)b=fx-acx+cx^{2}+ax^{2}-x^{3}$$

Divide both sides by ax-ac-x^{2}+xc.

$$\frac{\left(-x^{2}+ax+cx-ac\right)b}{-x^{2}+ax+cx-ac}=\frac{x\left(f-ac+cx+ax-x^{2}\right)}{-x^{2}+ax+cx-ac}$$

Dividing by ax-ac-x^{2}+xc undoes the multiplication by ax-ac-x^{2}+xc.

$$b=\frac{x\left(f-ac+cx+ax-x^{2}\right)}{-x^{2}+ax+cx-ac}$$

Divide x\left(-x^{2}+ax+cx-ac+f\right) by ax-ac-x^{2}+xc.

$$b=\frac{x\left(f-ac+cx+ax-x^{2}\right)}{\left(a-x\right)\left(x-c\right)}$$