Solve x^4/4-x^3+x^2+1+c=0

$$\frac{ { x }^{ 4 } }{ 4 } - { x }^{ 3 } + { x }^{ 2 } +1+c=0$$

Solve for c

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

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$x^{4}-4x^{3}+4x^{2}+4+4c=0$$

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

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

Add 4x^{3} to both sides.

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

Subtract 4x^{2} from both sides.

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

Subtract 4 from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

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