Let a be an integer such that all the real roots of the polynomial...

Let $2{x}^{5} + 5{x}^{4} + 10{x}^{3} + 10{x}^{2} + 10x + 10 = f\left(x\right)$

Now $f\left(- 2\right) = - 34 and$f\left(- 1\right) = 3$</p><p>Hence$f\left(x\right)$has a root in \left(- 2, - 1\right)$

Further ${f}^{'}\left(x\right) = 10{x}^{4} + 20{x}^{3} + 30{x}^{2} + 20x + 10$

$= 10{x}^{2}\left({x}^{2} + 2x + 3 + \dfrac{2}{x} + \dfrac{1}{{x}^{2}}\right)$

$= 10{x}^{2}\left[\left({x}^{2} + \dfrac{1}{{x}^{2}}\right) + 2\left(x + \dfrac{1}{x}\right) + 3\right]$

$= 10{x}^{2}\left[{\left(x + \dfrac{1}{x}\right)}^{2} - 2 + 2\left(x + \dfrac{1}{x}\right) + 3\right]$

$= 10{x}^{2}\left[{\left(x + \dfrac{1}{x} + 1\right)}^{2}\right] > 0$ for all $x$ belongs to $R$.

$f\left(x\right)$ is strictly increasing function. Since it is an odd degree polynomial it will have exactly one real root.

Hence, $f$\left(x\right)$has only one real root, so \left|a\right| = 2$.