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If P(x) be a polynomial satisfying the identity $P(x^{2}) +2x^{2} +10x

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If P(x) be a polynomial satisfying the identity $P(x^{2}) +2x^{2} +10x  [#permalink]

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New post 29 Jan 2019, 09:13
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If \(P(x)\) be a polynomial satisfying the identity \(P(x^{2}) +2x^{2} +10x = 2x\ P( x+1) +3\), then \(P(x)\) is
A) \(2x+3\)
B) \(3x-4\)
C) \(3x+2\)
D) \(2x–3\)
E) \(3x–3\)

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If P(x) be a polynomial satisfying the identity $P(x^{2}) +2x^{2} +10x  [#permalink]

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New post 29 Jan 2019, 09:51
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P(x) = ax + b => P(x+1) = a(x+1) + b
P(x^2) + 2x^2 + 10x = ax^2 + b + 2x^2 + 10x
2x*P(x+1) + 3 = 2x*[a(x+1) +b] + 3

P(x^2) + 2x^2 + 10x = 2x*P(x+1) + 3
<=> ax^2 + b + 2x^2 + 10x = 2x*[a(x+1) +b] + 3
<=> (a+2)*x^2 + 10x + b = 2a*x^2+ 2ax + 2bx + 3
<=> (a+2) * x^2 + 10x + b = 2a *x^2 + (2a+2b)*x + 3
we have a+2 = 2a => a=2
b is found by 2 ways: b=3 or 10 = 2a + 2b => b = (10-2*2)/2 = 3
=> P(x) = 2x + 3
A

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If P(x) be a polynomial satisfying the identity $P(x^{2}) +2x^{2} +10x  [#permalink]

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New post 26 Feb 2019, 04:17
P(x^2)+2x^2+10x=2x P(x+1)+3
=> P(x^2)+2x^2+10x-2xP(x+1)-3=0
the above expression carries only one constant "-3" , all the other terms are variables except the constant in the term "P(x^2)"
in order for the above expression to be zero the constant must be "+3"
which is only given in option 2x+3
OPTION A
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If P(x) be a polynomial satisfying the identity $P(x^{2}) +2x^{2} +10x   [#permalink] 26 Feb 2019, 04:17
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