P(x) = $x^{6}$^{ }− $5x^{4}$$+$ $16x^{2}$^{ }−$72x$ $+$ $9$
- $P(0) = 9$
- $P(-1)= -ve$
- $P(4) = +ve$
Hence, at least $2$ real roots can be clearly seen, but what about other $4$ roots left ?
That's why we check $P^{n}\left ( x \right )$,
$P'(x)$ = $6x^{5}$^{ }- $20x^{3}$^{ }+ $32x$$-72$
$P''(x)$ $=$ $30x^{4}$ - $60x^{2}$$+32>0$ for any real value of x.
comparing this with a quadratic eq taking $x^{2}$as y we get $30y^{2}$$-60y+32$. The discriminant $\left ( b^{2}-4ac \right )$ is negative implying $P''(x)$ has no real roots
Hence by Rolle's theorem $P'(x)$ can have at most $1$ real root and $P(x)$ can have at most $2$ real roots. Because if a function $f(x)$ has $2$ roots $x_{1}$ and $x_{2}$ then there exists a point $x ∈$ $\left [ x1,x2 \right ]$ where the curve becomes flat i.e its the root of $f'(x)$ meaning $f(x)$ has max $2$ roots
=> Exactly two distinct real roots
Option A