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(x^2+1)(x+1)

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Carcass
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(x^2+1)(x+1) [#permalink] New post   05 Jan 2020, 03:36
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\(x > 0\)


Quantity A
Quantity B
\((x^2+1)\)\((x+1)\)
\((x+1)^2(x-1)\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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Re: (x^2+1)(x+1) [#permalink] New post   05 Jan 2020, 03:45
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Carcass wrote:
\(x > 0\)


Quantity A
Quantity B
\((x^2+1)\)\((x+1)\)
\((x+1)^2(x-1)\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Kudos for the right answer and explanation


QTY A::

\((x^2+1)\)\((x+1)\)

QTY B::

\((x+1)^2(x-1)\) = \((x+1)(x+1)(x-1)\)

REMOVING x+1 from both quantities

QTY A : \((x^2+1)\)

QTY B : \((x+1)(x-1)\) = \((x^2-1) \)

Therefore QTY A > QTY B
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void
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(x^2+1)(x+1) [#permalink] New post   27 Aug 2021, 01:24
But we can divide (x-1) both side,then
Qt. A = (x+1)(x+1)

Qt. B = (x+1)(x+1)

ans-C.
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Re: (x^2+1)(x+1) [#permalink] New post   27 Aug 2021, 02:21
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Sir,

How to divide \((x^2+1)(x+1)\) by \(x-1\) ?

void wrote:
But we can divide (x-1) both side,then
Qt. A = (x+1)(x+1)

Qt. B = (x+1)(x+1)

ans-C.
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void
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Re: (x^2+1)(x+1) [#permalink] New post   27 Aug 2021, 02:44
rx10 wrote:
Sir,

How to divide \((x^2+1)(x+1)\) by \(x-1\) ?

void wrote:
But we can divide (x-1) both side,then
Qt. A = (x+1)(x+1)

Qt. B = (x+1)(x+1)

ans-C.

(x^2+1) follows A^2+B^2=(A-B)(A+B)
(x^2+1^2)=(x-1)(x+1)
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rx10
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(x^2+1)(x+1) [#permalink] New post   27 Aug 2021, 03:07
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a^2 - b^2 = (a + b) (a - b)
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