\(f(x)=x^4+29x^2+100=\left(x^2\right)^2+29x^2+100\\\\x^2=v;\ v\geq0\\\\f(v)=v^2+29v+100\\\\a=1;\ b=29;\ c=100\\\\\Delta=b^2-4ac;\ v_1=\frac{-b-\sqrt\Delta}{2a};\ v_2=\frac{-b+\sqrt\Delta}{2a}\\\\\Delta=29^2-4\cdot1\cdot100=841-400=441;\ \sqrt\Delta=\sqrt{441}=21\\\\v_1=\frac{-29-21}{2\cdot1}=\frac{-40}{2}=-20 < 0;\ v_2=\frac{-29+21}{2\cdot1}=\frac{-8}{2}=-4 < 0\)
\(Function\ has\ no\ zeros.\\\\\\f(x)=x^4+29x^2+100=x^4+25x^2+4x^2+100\\\\=x^2(x^2+25)+4(x^2+25)=(x^2+25)(x^2+4)\\\\is\ not\ impossible\ write\ the\ polynomia\l as\ a\ product\ of\ linear\ factors.\)
