3 octagons 4 Pentagons 2 trapezoids how many sides are on the shapes altogether
1 answer:
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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
Explanation:
Let's solve it smartly!
Given expression:
Factors: (x+p)(x+q)
Condition: c<0
Now let us expand (x+p)(x+q):
=>
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