Rational Zeroes

Algebra Level 2

Find the sum of all the rational roots of the equation

x 5 − 4 x 4 + 2 x 3 + 2 x 2 + x + 6 = 0. x^5-4x^4+2x^3+2x^2+x+6=0.


The answer is 4.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Rishabh Jain
Jun 15, 2016

Call f ( x ) = x 5 − 4 x 4 + 2 x 3 + 2 x 2 + x + 6 f(x)=x^5-4x^4+2x^3+2x^2+x+6 .

By rational root theorem we need to check among ± 1 , ± 6 , ± 2 , ± 3 \pm1,\pm 6,\pm 2,\pm 3 . By simple substitution we see x = − 1 , 2 , 3 x=-1,2,3 are required solutions . Hence ( x + 1 ) ( x − 2 ) ( x − 3 ) (x+1)(x-2)(x-3) is a factor of given equation. Dividing by this:-

f ( x ) = ( ( x + 1 ) ( x − 2 ) ( x − 3 ) ) ( x 2 + 1 ) = 0 f(x)=((x+1)(x-2)(x-3))(x^2+1)=0

The fourth bracket gives imaginary roots in pair while only rational solutions are − 1 , 2 , 3 -1,2,3 whose sum is 4 \large\boxed 4 .

19 Helpful 0 Interesting 0 Brilliant 0 Confused

hello,the answer can be modified by using the vieta formulae

abhishek alva - 4 years, 8 months ago

The complex roots should have had real parts, Vieta's formula is working :3

Pratyush Pandey - 4 years, 4 months ago

i used sum of roots=

-b/a

Kushagra Kanchan - 1 year, 10 months ago

Sorry, you have to correct the result -1+2+3= 4. Cheers :D

Azmi Muhammad - 4 years, 11 months ago

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Yep... A typo..... Corrected

Rishabh Jain - 4 years, 11 months ago
Ariijit Dey
Jul 14, 2018

I used Vieta's Theorem to find the sum

2 Helpful 0 Interesting 0 Brilliant 0 Confused

In that case, it only works because the real part of the complex numbers is 0 (i - i = 0).

Nicolas Kevin - 2 years, 7 months ago

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