Suppose you can factor x^2 +bx +c as (x+p)(x+q) . if c< 0, what could be the possible values of p and q?
a: p= -3, q= 7
b: p= 11, q= 4
c: p= -2, q= -5
d: p= 1, q= 10

Ans: Option A

Explanation:
Let's solve it smartly!
Given expression:  --- (A)
Factors: (x+p)(x+q)
Condition: c<0
Now let us expand (x+p)(x+q):
=>   --- (B)

By comparing (B) with (A), we can say that:
pq = c --- (C)

Now, as the condition says, c<0, it means either p or q is negative. Both cannot be positive or both cannot be negative.

1) If p>0, q>0, it means c>0 since (+p)(+q) = (+c)(according to equation (C)). Condition is not met.
Hence, option B and D are wrong.

2) If p<0, q<0 it means c>=0 since (-p)(-q) = (+c)(according to equation (C)). Condition is not met.
Hence option C is out as well.

We are left with Option A:p<0, q>0 it means c<0 since (-p)(+q) = (-c)(according to equation (C)).
Condition is MET!
Hence,
Ans: Option A: p= -3, q= 7

Ithink the answer is discovery verbs because they are quite similar to the verb 'discovering'