Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
$$x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = (x - 1)(x - 2)(x - 3)(x - 4)$$ then
$$(x - 1)(x - 4) = x^2 - 5x + 4$$ and $$(x - 2)(x - 3) = x^2 - 5x + 6$$. With $$t = x^2 - 5x$$, the product is $$(t + 4)(t + 6) = t^2 + 10t + 24 = x^4 - 10x^3 + 35x^2 - 50x + 24$$. So $$a_3 = -10$$, $$a_2 = 35$$, $$a_1 = -50$$, $$a_0 = 24$$, and all four options a, b, c and d are correct.
Was this solution helpful?
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.