Cubic polynomial

[stuart clark]

Let a,b,c,d be four integers such that ad is odd and bc is even,then ax³+bx²+cx+d = 0 has

options::

(a) at least  one irrational roots
(b) all three rational roots
(c) all three integral roots
(d) none of these

[senior]

think about a = 1, b = 2, c = 2, d = 1
Then P(x) = x3 + 2x² + 2x + 1 = x²(x+1) + (x+1)² = (x+1)(x²+x+1)
which has roots x = -1, and two imaginary ones.
This holds with the condition but with none of choices a,b, or c. So, choice d.

[stuart clark]

Thanks Moderator Got it