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20 Apr 2026, 00:58
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Difficulty:
55%
(hard)
Question Stats:
56% (02:46) correct
44%
(02:03)
wrong
based on 25
sessions
History
Date
Time
Result
Not Attempted Yet
If two factors of \(2x^3 - px + q\) are \(x + 2\) and \(x - 1\), what is the value of \(p + q\)?
(A) 6
(B) 8
(C) 9
(D) 10
(E) 12
(A) 6
(B) 8
(C) 9
(D) 10
(E) 12
20 Apr 2026, 07:11
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x=-2
x=1
So plug them in.
From x=−2:
2(−2)^3−p(−2)+q=0
2(−8)+2p+q=0
−16+2p+q=0
2p+q=16
From x=1:
2(1)^3−p(1)+q=0
2−p+q=0
−p+q=−2
So the system is:
1)2p+q=16
2)−p+q=−2
q=p-2
Substitue in 1):
2p+p-2=16
Solve for p
p=6
Then:
q=6-2 --> q=4
Thus:
p+q=10
x=1
So plug them in.
From x=−2:
2(−2)^3−p(−2)+q=0
2(−8)+2p+q=0
−16+2p+q=0
2p+q=16
From x=1:
2(1)^3−p(1)+q=0
2−p+q=0
−p+q=−2
So the system is:
1)2p+q=16
2)−p+q=−2
q=p-2
Substitue in 1):
2p+p-2=16
Solve for p
p=6
Then:
q=6-2 --> q=4
Thus:
p+q=10
20 Apr 2026, 07:11
Kudos
Bookmarks
x=-2
x=1
So plug them in.
From x=−2:
2(−2)^3−p(−2)+q=0
2(−8)+2p+q=0
−16+2p+q=0
2p+q=16
From x=1:
2(1)^3−p(1)+q=0
2−p+q=0
−p+q=−2
So the system is:
1)2p+q=16
2)−p+q=−2
q=p-2
Substitue in 1):
2p+p-2=16
Solve for p
p=6
Then:
q=6-2 --> q=4
Thus:
p+q=10
x=1
So plug them in.
From x=−2:
2(−2)^3−p(−2)+q=0
2(−8)+2p+q=0
−16+2p+q=0
2p+q=16
From x=1:
2(1)^3−p(1)+q=0
2−p+q=0
−p+q=−2
So the system is:
1)2p+q=16
2)−p+q=−2
q=p-2
Substitue in 1):
2p+p-2=16
Solve for p
p=6
Then:
q=6-2 --> q=4
Thus:
p+q=10
