Find sum of all possible integer(s) m such that the equation $x^{3} + (m + 1) x^{2} - (2 m - 1) x - (2 m^{2} + m + 4) = 0$ has an integer root.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 10.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Note that the left-hand side of the equation can be written as
$x (x^{2} - 2 m + 1) + m (x^{2} - 2 m + 1) + x^{2} - 2 m + 1 = 5$
Then, $(x + m + 1) (x^{2} - 2 m + 1) = 5$
Now, if the equation has an integer root, we have
$(x + m + 1, x^{2} - 2 m + 1) \in {(1, 5), (- 1, - 5), (5, 1), (- 5, - 1)}$
We have four cases,
(i) $x + m + 1 = 1 a n d x^{2} - 2 m + 1 = 5.$
Then, $x^{2} - 2 m + 1 + 2 (x + m + 1) = 7.$
Hence, $x^{2} + 2 x - 4 = 0,$ which has no integer roots.
(ii) $x + m + 1 = - 1 a n d x^{2} - 2 m + 1 = - 5.$ Then, $x^{2} + 2 x + 10 = 0,$ which has no real roots.
(iii) $x + m + 1 = 5 a n d x^{2} - 2 m + 1 = 1.$ Then $x^{2} + 2 x - 8 = 0,$ which gives $x = 2, - 4 a n d, m = 2, 8$ respectively.
(iv) $x + m + 1 = - 5 a n d x^{2} - 2 m + 1 = - 1.$ Then $x^{2} + 2 x + 14 = 0,$ which has no real roots.