Hey Vieta :D

Algebra Level 4

P ( x ) = ( x − 2 ) ( x 5 + 2 x 4 + x 3 + x 2 + 4 x + 1 ) + ( x − 4 ) ( x 5 + 2 x 4 + x 3 + x 2 + 4 x + 1 ) P(x) = (x-2)(x^5+2x^4+x^3+x^2+4x+1)+(x-4)(x^5+2x^4+x^3+x^2+4x+1)

What is the sum of the roots of P ( x ) = 0 P(x) = 0 ?

3 1 7 \sqrt{7} 2

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Paul Ryan Longhas
Feb 20, 2015

Factor P ( x ) P(x) to get P ( x ) = ( 2 x − 6 ) ( x 5 + 2 x 4 + x 3 + x 2 + 4 x + 1 ) P(x) = (2x-6)(x^5+2x^4+x^3+x^2+4x+1) which sum of the roots is 3 − 2 = 1 3 - 2 = 1

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Could u post a detailed sol

Naman Kapoor - 5 years, 6 months ago

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P ( x ) = ( 2 x − 6 ) ⏟ x = 3 ( x 5 + 2 x 4 + x 3 + x 2 + 4 x + 1 ) ⏟ ∑ x = − 2 ⏟ ∑ x = 3 − 2 = 1 P(x) = \underbrace{\underbrace{(2x-6)}_{ x=3} \underbrace{(x^5+2x^4+x^3+x^2+4x+1)}_{\sum x =-2}}_{\sum x= 3-2=1}

Akshat Sharda - 5 years, 4 months ago

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can you please tell why have you added the sum of roots in both cases .. please elaborate

Deepansh Jindal - 5 years, 1 month ago

Exactly !! This was an easy problem !

Akshat Sharda - 5 years, 4 months ago

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