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If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x =

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Bunuel
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If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   25 Nov 2021, 03:30
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If \(P(x)\) be a polynomial satisfying the identity \(P(x^2) +2x^2 +10x = 2xP(x+1) +3\), then \(P(x)\) is

(A) \(2x+3\)

(B) \(3x-4\)

(C) \(3x+2\)

(D) \(2x–3\)

(E) \(3x–3\)
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A
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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   02 Dec 2021, 01:41
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Looking at both the left and the right side one sees that there is a constant of positive 3 on the right side. For these the two sides to be equal there needs to be a balance. Looking at the left side, the only way that there will be a constant of 3 is if the function itself contains positive 3. The only answer choice which has this is A.

Testing answer choice A: Plugging in \(2x + 3\) into \(P(x^2) + 2x^2 + 10x = 2xP(x+1) + 3\)

\(2x^2 + 3 + 2x^2 + 10x = 2x*[2(x+1) + 3] + 3\)

\(4x^2 + 10x + 3 = 2x*[2x + 2 + 3] + 3\)

\(4x^2 + 10x + 3 = 4x^2 + 10x +3\)

Both sides are equal to one another.

Answer A
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If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   02 Dec 2021, 05:13
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A look at the answer choices confirms that the answer is of the form ax + b.
So, let us assume that P(x) = ax + b.
Thus
P(x) = ax + b
\(P(x^2) = ax^2 + b\)
\(P(x+1) = ax + a + b\)

Substitute these 3 statements in the question.

\(ax^2 + 2x^2 + 10x + b\) = \(2ax^2 + 2ax +2bx + 3\)

This gives us b = 3, and a = 2.
Thus Option a is the answer
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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   02 Dec 2021, 08:12
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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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   02 Dec 2021, 08:36
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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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   18 Dec 2021, 07:20
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Kinshook wrote:
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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   18 Dec 2021, 07:34
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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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   28 Oct 2023, 23:52
I am really getting confused how to plug in the values. IF some one can help, it will be great
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If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   30 Oct 2023, 12:22
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DrAnkita91 wrote:
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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink] New post   31 Oct 2023, 05:50
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youcouldsailaway wrote:
DrAnkita91 wrote:
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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Re: If P(x) be a polynomial satisfying the identity P(x^2) + 2x^2 + 10x = [#permalink]
31 Oct 2023, 05:50
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