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

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

Solve for p

$$\left\{\begin{matrix}\\p=x^{3}-3x^{2}+x+4\text{, }&\text{unconditionally}\\p\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 1 from both sides.

$$-px=-x^{4}+3x^{3}-x^{2}-4x+1-1$$

Subtract 1 from 1 to get 0.

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

The equation is in standard form.

$$\left(-x\right)p=-x^{4}+3x^{3}-x^{2}-4x$$

Divide both sides by -x.

$$\frac{\left(-x\right)p}{-x}=\frac{x\left(-x^{3}+3x^{2}-x-4\right)}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$p=\frac{x\left(-x^{3}+3x^{2}-x-4\right)}{-x}$$

Divide x\left(-x^{3}+3x^{2}-x-4\right) by -x.

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