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Bunuel
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Any other approach to the sum? I woild have never thought to do e/a!

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Bunuel
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


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Putting x=5 in \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\)

IMO C­
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Bunuel
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9

Are You Up For the Challenge: 700 Level Questions


Thanks shameekv1989

due to some technical problems in which I cannot post some letters correctly to the borad, i put it in the attached file
i write down some way of thinking and related theorem on the board­
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If 2x^5 − 14x^4 + 31x^3.docx [27.57 KiB]
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mimishyu

Bunuel
If \(2x^5 − 14x^4 + 31x^3 − 64x^2 + 19x + 130 = 0\), then which of the following could be the value of x ?

A. 1
B. 3
C. 5
D. 7
E. 9


Are You Up For the Challenge: 700 Level Questions


Thanks shameekv1989

due to some technical problems in which I cannot post some letters correctly to the borad, i put it in the attached file
i write down some way of thinking and related theorem on the board
Hey mimishyu - Amazing additional information. Thanks for confirming the approach.­
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Thank you for your appreciation,I translated all from Chinese website since I cannot find any related material in English which can better help to explain this ques, before I post I'm quite nervous whether I will do sth wrong

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2x^5−14x^4+31x^3−64x^2+19^x+130=0

product of all = -130*2= -260
Sum of all = 19/2
Now we can either go to each option and evaluate the expression
Or we could go for smart guess.

If the product is -260 then assuming that at least one integer will be there
We can go for sure that 10 will be multiple.
So from options 1 and 5 seems the right options.
Keep x = 1 and the equation won't hold.

So x = 5
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