2026/041) For $x^2+x+5$ to be a factor of $x^4+px^2+q$ the values of p and q must be, respectively: (A) −2,5(B) 5,25(C) 10,20(D) 6,25(E) 14,25

 Because product is bi-quadratic and one factor is quadratic so other factor must be quadratic

Other factor is of the form $x^2+ax+b$

So we get 

 $(x^2+2x+5)(x^2+ax+b) = = x^4 + (2+a)x^3 + (b+2a+5)x^2 + (2b+5a)x + 5b$

Comparing with $x^4+px^2+ q$ we get a = - 2 (coefficient $x^3$) and $b = 5$ from coefficient of x

So q = 5b = 25

Comparing coefficient of x^2 we get p = 5 -4 + 5 = 6

So Ans is (D) 6,25