Solve x^4+ax^3-bx^2-8x+12=0

$$x ^ { 4 } + a x ^ { 3 } - b x ^ { 2 } - 8 x + 12 = 0$$

Solve for a

$$a=\frac{-x^{4}+bx^{2}+8x-12}{x^{3}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Add 8x to both sides. Anything plus zero gives itself.

$$x^{4}+ax^{3}-bx^{2}+12=8x$$

Subtract 12 from both sides.

$$x^{4}+ax^{3}-bx^{2}=8x-12$$

Add bx^{2} to both sides.

$$x^{4}+ax^{3}=8x-12+bx^{2}$$

Subtract x^{4} from both sides.

$$ax^{3}=8x-12+bx^{2}-x^{4}$$

The equation is in standard form.

$$x^{3}a=-x^{4}+bx^{2}+8x-12$$

Divide both sides by x^{3}.

$$\frac{x^{3}a}{x^{3}}=\frac{-x^{4}+bx^{2}+8x-12}{x^{3}}$$

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

$$a=\frac{-x^{4}+bx^{2}+8x-12}{x^{3}}$$

Divide 8x-12+bx^{2}-x^{4} by x^{3}.

$$a=\frac{bx^{2}+8x-12}{x^{3}}-x$$

Solve for b

$$b=\frac{x^{4}+ax^{3}-8x+12}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Add 8x to both sides. Anything plus zero gives itself.

$$x^{4}+ax^{3}-bx^{2}+12=8x$$

Subtract 12 from both sides.

$$x^{4}+ax^{3}-bx^{2}=8x-12$$

Subtract x^{4} from both sides.

$$ax^{3}-bx^{2}=8x-12-x^{4}$$

Subtract ax^{3} from both sides.

$$-bx^{2}=8x-12-x^{4}-ax^{3}$$

The equation is in standard form.

$$\left(-x^{2}\right)b=-x^{4}-ax^{3}+8x-12$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)b}{-x^{2}}=\frac{-x^{4}-ax^{3}+8x-12}{-x^{2}}$$

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

$$b=\frac{-x^{4}-ax^{3}+8x-12}{-x^{2}}$$

Divide 8x-12-x^{4}-ax^{3} by -x^{2}.

$$b=ax+x^{2}+\frac{12-8x}{x^{2}}$$