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Bunuel
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Quickest way is to look at the integer on the right hand side and it can only be cancelled through P(x) function, the only option with +3 is option A

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Asked: If \(P(x)\) be a polynomial satisfying the identity \(P(x^2) +2x^2 +10x = 2xP(x+1) +3\), then \(P(x)\) is

\(P(x^2) +2x^2 +10x = 2xP(x+1) +3\)

(A) \(2x+3\)
P(x^2) +2x^2 +10x = 2x^2 + 3 + 2x^2 + 10x = 4x^2 + 10x + 3
2xP(x+1) +3 = 2x(2x + 5) + 3 = 4x^2 + 10x + 3

(B) \(3x-4\)

(C) \(3x+2\)

(D) \(2x–3\)

(E) \(3x–3\)

IMO A
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Kinshook
Asked: If \(P(x)\) be a polynomial satisfying the identity \(P(x^2) +2x^2 +10x = 2xP(x+1) +3\), then \(P(x)\) is

\(P(x^2) +2x^2 +10x = 2xP(x+1) +3\)

(A) \(2x+3\)
P(x^2) +2x^2 +10x = 2x^2 + 3 + 2x^2 + 10x = 4x^2 + 10x + 3
2xP(x+1) +3 = 2x(2x + 5) + 3 = 4x^2 + 10x + 3

(B) \(3x-4\)

(C) \(3x+2\)

(D) \(2x–3\)

(E) \(3x–3\)

IMO A

I did not understand this question. How are we plugging in the values here? Could you please explain me in detail? Kinshook
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The easiest way would be substitute x=0, the whole equation becomes P(0)=3, and compare the option and then substitute x=0 in options, the answer is A.
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I am really getting confused how to plug in the values. IF some one can help, it will be great
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DrAnkita91
I am really getting confused how to plug in the values. IF some one can help, it will be great

If you take \(2x +3\) (which is op. A), you would substitute \(x \) with \(x^2\) because its \(P(x^2)\) on the LHS and then on RHS, you would substitute \(x\) with \(x+1\) because its \(P(x+1)\) on the RHS.

However, you'll also add the constant term (3 in this case) as the polynomial \(2x + 3\) is what P stands for.

In short wherever you see P, substitute one of the answer choices but make \(x\) as per the \(x\) in the question stem, i.e, in this case, \(x^2\) and \(x + 1\).
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youcouldsailaway
DrAnkita91
I am really getting confused how to plug in the values. IF some one can help, it will be great

If you take \(2x +3\) (which is op. A), you would substitute \(x \) with \(x^2\) because its \(P(x^2)\) on the LHS and then on RHS, you would substitute \(x\) with \(x+1\) because its \(P(x+1)\) on the RHS.

However, you'll also add the constant term (3 in this case) as the polynomial \(2x + 3\) is what P stands for.

In short wherever you see P, substitute one of the answer choices but make \(x\) as per the \(x\) in the question stem, i.e, in this case, \(x^2\) and \(x + 1\).


Thank you. I understood
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