If 2x^5 14x^4 + 31x^3 64x^2 + 19x + 130 = 0, then which of the fol

Question

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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shameekv1989 · Accepted Answer

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

The product of the roots =  e/a  = 130/2 = 65 = 13*5 = Product of 2 co primes.

Thus one of the roots should be 5 from the given options IMO

Answer - C

lacktutor · Answer

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

2x^{5}−14x^{4}+31x^{3}−64x^{2}+19x+130=

(2x^{5}- 10x^{4}) -(4x^{4}- 20x^{3}) +(11x^{3} - 55x^{2}) -(9x^{2} -45x) -(26x -5*26)=

2x^{4}( x-5) -4x^{3}(x-5)+ 11x^{2}(x-5 ) -9x(x-5) -26(x-5 )=

( x-5)*(2x^{4}- 4x^{3}+ 11x^{2}-9x -26 ) =0

-->  x  could be  5 .

Answer (C)

MS2020 · Answer

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

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Kinshook · Answer

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

IMO C

mimishyu · Answer

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

shameekv1989 · Answer

Hey  mimishyu  - Amazing additional information. Thanks for confirming the approach.

mimishyu · Answer

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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sambitspm · Answer

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

prayforgigi · Answer

I agree with all previous responses, however, I think there's a faster (though less rigorous) shortcut to solve this problem.

We can factor out  x   and rewrite the equation as:

x*(2x^4 - 14x^3 + 31x^2 -64x +19) = -130

Since x is an integer and all coefficients are integers, then what is inside the parenthesis must be an integer. Let's call that integer   A   . Thus:

x*(A) = -130

From this, we get 130 must be a multiple of x. Neither 3,7 or 9 are divisors of 130, we can discard those. Only possible answers are  x = 5  or  x = 1

We plug in  x = 1  to check, we can quickly see that   2-14+31-64+19 < 130 . Thus we can discard   x = 1.

Only possible answer is  x = 5

Answer: C

If 2x^5 14x^4 + 31x^3 64x^2 + 19x + 130 = 0, then which of the fol

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If 2x^5 14x^4 + 31x^3 64x^2 + 19x + 130 = 0, then which of the fol

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards