Question

Let $P\left(x\right)=x^3-a{x}^2+bx+c$ where $a,b,c\in R$ has integral roots such that $P\left(6\right)=3,$ then number of positive integral divisors of sum of possible value of $a$ is

Answer 1

$P\left(x\right)=(x-\alpha )(x-\beta )(x-\gamma )$
Substituting $x=6$, we get
$3=(6-\alpha )(6-\beta )(6-\gamma )$
Now, possible combinations of roots satisfying the above equation
$\left(i\right)\alpha =3,\beta =5,\gamma =5\Rightarrow a=13$
$\left(ii\right)\alpha =3,\beta =7,\gamma =7\Rightarrow a=17$
$\left(iii\right)\alpha =9,\beta =5,\gamma =7\Rightarrow a=21$
Now, sum of all values of $a$ is $51=17\times 3$
Total number of positive integral divisors of 51 $=4$