Verified answer

[tex]\displaystyle\bf\\f(x)=4x^{3} +4x\\\\F(x)=4\cdot \frac{x^{4} }4} +4\cdot \frac{x^{2} }{2} +C\\\\\\F(x)=x^{4} +2x^{2} +C\\\\F(x)=0 \ \ ; \ \ x=-1\\\\0=(-1)^{4} +2\cdot (-1)^{2} +C\\\\1+2\cdot 1+C=0\\\\C=-3\\\\\boxed{F(x)=x^{4} +2x^{2} -3}\\\\\\x^{4} +2x^{2} -3=0\\\\(x^{4} -x^{2} )+(3x^{2} -3)=0\\\\x^{2} \cdot(x^{2} -1)+3\cdot(x^{2} -1)=0\\\\(x^{2} -1)\cdot(x^{2} +3)=0\\\\(x-1)(x+1)(x^{2} +3)=0[/tex]

[tex]\displaystyle\bf\\x^{2} +3\neq 0 \ \ \ \Rightarrow \ \ \ \left[\begin{array}{ccc}x-1=0\\x+1=0\end{array}\right \ \ \ \Rightarrow \ \ \ \left[\begin{array}{ccc}x=1\\x=-1\end{array}\right[/tex]

Ещё один нуль первообразной равен 1 .