Solve 5*f(x)=-(x+3)^4+1

$$5 \cdot f ( x ) = - ( x + 3 ) ^ { 4 } + 1$$

Solve for f

$$f=-\frac{\left(x+2\right)\left(x+4\right)\left(x^{2}+6x+10\right)}{5x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

The equation is in standard form.

$$5xf=-\left(x+3\right)^{4}+1$$

Divide both sides by 5x.

$$\frac{5xf}{5x}=-\frac{\left(x+2\right)\left(x+4\right)\left(\left(x+3\right)^{2}+1\right)}{5x}$$

Dividing by 5x undoes the multiplication by 5x.

$$f=-\frac{\left(x+2\right)\left(x+4\right)\left(\left(x+3\right)^{2}+1\right)}{5x}$$

Divide -\left(\left(x+3\right)^{2}+1\right)\left(x+4\right)\left(x+2\right) by 5x.

$$f=-\frac{x^{3}}{5}-\frac{12x^{2}}{5}-\frac{54x}{5}-\frac{108}{5}-\frac{16}{x}$$