Fundamental of Mathematics
If P ( x ) is a cubic polynomial with leading coefficient 2 such that P ( 1 ) = 2 , P ( 2 ) = 8 , P ( 3 ) = 1 8 , find P ( 4 ) .
The answer is 44.
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2 solutions
Thank you..
Let, f ( x ) = P ( x ) − 2 x 3
Now we can solve it using Method of Differences ,
x 1 2 3 4 f ( x ) 0 − 8 − 3 6 − 8 4 D 1 ( x ) − 8 − 2 8 − 4 8 D 2 ( x ) − 2 0 − 2 0
Therefore,
∴ P ( x ) P ( 4 ) = f ( x ) + 2 x 3 = f ( 4 ) + 2 ( 4 ) 3 = − 8 4 + 1 2 8 = 4 4
Shouldn't it be 2 x 2 ?
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What's the reason for thinking that it should be 2 x 2 ?
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Ah, my bad, I read something wrong the first time. Your method was to turn the cubic into a quadratic. I confused it with the other method in which we could have put f ( x ) = 2 ( x − 1 ) ( x − 2 ) ( x − 3 ) + 2 x 2
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@A Former Brilliant Member – No problem !! Be careful !!
Polynomial ,P(x) = K [ (x-1)(x-2)(x-3) ]+ 2 x 2
Given: Leading coefficient is 2, so K=2
P(4) = 2[ (3)(2)(1) ]+2[ (4)(4) ]
P(4) = 12+32
P(4) = 44