Polynomials Again
Let p ( x ) be a fourth degree monic polynomial such that p ( − 1 ) = 1 , p ( 2 ) = 4 , p ( − 3 ) = 9 , p ( 4 ) = 1 6 . Calculate p ( 1 )
The answer is 25.
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2 solutions
The most important point that needs to be mentioned here is the usage of Remainder-Factor Theorem for obtaining the polynomial p ( x ) .
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Thanks for mentioning that. But How should I include it in my solution?
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First, you note that,
p ( x ) = x 2 ∀ x ∈ { − 1 , 2 , − 3 , 4 }
Next, you consider a polynomial g which is of the same degree as p and is also monic defined by g ( x ) = p ( x ) − x 2 .
So, you have g ( x ) = 0 ∀ x ∈ { − 1 , 2 , − 3 , 4 } . By Remainder-Factor theorem,
g ( x ) = ( x + 1 ) ( x − 2 ) ( x + 3 ) ( x − 4 ) ⟺ p ( x ) = ( x + 1 ) ( x − 2 ) ( x + 3 ) ( x − 4 ) + x 2
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@Prasun Biswas – Ahh, Great. But I have to go somewhere now, I will add it as soon as I come back :)
@Prasun Biswas , I don't have the faintest idea as to why people downvote your comments :3
Good solution .Up voted.
nice solution by Mehul. But the solution is particular where p(x)= x^2 for all x belonging to the set {-1,2,-3,4}is vividly satisfied. However, this can not be a general solution for all monic polynomials of degree four. For general solution, let p(x)= x^4+ax^3+bx^2+cx+d. Next obtain four simultaneous equations in four variables a,b.c and d like p(-1)=1-a+b-c+d=1 and three other similar equations as per given conditions. Solving these four equations, we obtain a=-2, b=-12, c=14 and d=24. Hence, the polynomial becomes p(x)= x^4-2x^3-12x^2+14x+24. Plug in x=1 and get p(1)= 25
From the given values of the polynomial, We can say that:-
p ( x ) = ( x + 1 ) ( x − 2 ) ( x + 3 ) ( x − 4 ) + x 2
Now, Plugging in the value of 1,
p ( 1 ) = ( 1 + 1 ) ( 1 − 2 ) ( 1 + 3 ) ( 1 − 4 ) + 1
= 2 × ( − 1 ) × 4 × ( − 3 ) + 1 = 2 5
Hence the answer is 25
QED. Cheers!