(i) $f(x)=x^{4}+4x+c$
$f'(x)=4x^{3}+4=0$ $\Rightarrow x^{3}=-1$ or $x=-1$
$f''(x=-1)=12> 0$ // this shows that function has only one minima at $x=-1$.
this shows that above function will somewhat look like upper parabola, therfore it can have atmost 2 roots only.
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(ii) $f(x)=x^{5}+10x+3$
$f'(x)=5x^{4}+10>0(always)$
since it is monotonically increasing function, it will cut x axis only once, therefore it will have only 1 root.