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What is the product of all the roots of the following equation?

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What is the product of all the roots of the following equation?  [#permalink]

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New post Updated on: 22 Aug 2017, 21:00
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What is the product of all the roots of the following equation?

\(x^5 - 3x^4 = 3x^2 - 3x^3 + 2x\)

A. 28
B. 19
C. 14
D. 1
E. 0
Spoiler: OA

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Originally posted by ChiranjeevSingh on 22 Aug 2017, 19:49.
Last edited by ChiranjeevSingh on 22 Aug 2017, 21:00, edited 5 times in total.
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What is the product of all the roots of the following equation?  [#permalink]

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New post 22 Aug 2017, 20:14
1
ChiranjeevSingh wrote:
What is the product of all the roots of the following equation?

\(x^5 - 3x^4 = 3x^2 - 3x^3 + 2x\)

A. 28
B. 14
C. 19
D. 0
E. 1

I'll post the OA in a while!



Hi..

X is common to all terms so one value of x is 0..
No need to go any further..
Product of all terms with one of them as 0 will be 0
E
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Re: What is the product of all the roots of the following equation?  [#permalink]

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New post 22 Aug 2017, 20:55
1
ChiranjeevSingh wrote:
What is the product of all the roots of the following equation?

\(x^5 - 3x^4 = 3x^2 - 3x^3 + 2x\)

A. 28
B. 14
C. 19
D. 0
E. 1

I'll post the OA in a while!


You have to post the OA from the beginning. Also, on the GMAT, answer choices are in ascending or descending order.
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Re: What is the product of all the roots of the following equation?  [#permalink]

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New post 22 Aug 2017, 23:31
ChiranjeevSingh wrote:
What is the product of all the roots of the following equation?

\(x^5 - 3x^4 = 3x^2 - 3x^3 + 2x\)

A. 28
B. 19
C. 14
D. 1
E. 0


since equation becomes
\(x(x^4-3x^3+3x^2-3x-2)=0\)
one of the roots will be x=0

Hence product of all roots will be 0
E
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Re: What is the product of all the roots of the following equation?  [#permalink]

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New post 23 Aug 2017, 03:07
Taking x common and all the terms on left side ,
x(x4−3x3+3x2−3x−2)=0
Which equuals ,

x=0 or x4-3x3=3x2-2=0

One of the root is 0 . That means product of roots = 0
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Re: What is the product of all the roots of the following equation?   [#permalink] 23 Aug 2017, 03:07
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