Solve (x)=x^4-2x^3-x^2+x-1p

$$( x ) = x ^ { 4 } - 2 x ^ { 3 } - x ^ { 2 } + x - 1 p$$

Solve for p

$$p=x^{2}\left(x^{2}-2x-1\right)$$

Steps for Solving Linear Equation

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

$$x^{4}-2x^{3}-x^{2}+x-p=x$$

Subtract x^{4} from both sides.

$$-2x^{3}-x^{2}+x-p=x-x^{4}$$

Add 2x^{3} to both sides.

$$-x^{2}+x-p=x-x^{4}+2x^{3}$$

Add x^{2} to both sides.

$$x-p=x-x^{4}+2x^{3}+x^{2}$$

Subtract x from both sides.

$$-p=x-x^{4}+2x^{3}+x^{2}-x$$

Combine x and -x to get 0.

$$-p=-x^{4}+2x^{3}+x^{2}$$

The equation is in standard form.

$$-p=x^{2}+2x^{3}-x^{4}$$

Divide both sides by -1.

$$\frac{-p}{-1}=\frac{x^{2}\left(1+2x-x^{2}\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$p=\frac{x^{2}\left(1+2x-x^{2}\right)}{-1}$$

Divide x^{2}\left(-x^{2}+2x+1\right) by -1.

$$p=x^{4}-2x^{3}-x^{2}$$