Roots of a polynomial

Algebra Level 4

How many possible non-real complex solution can given polynomial have ?

a x 5 + b x 4 + c x 3 + d x 2 + e x + f ax^5 + bx^4 + cx^3 + dx^2 + ex + f
where a , b , c , d , e , f \textit{a , b , c , d , e , f} are non-zero real constants.

0,1,5 0,2,4 0,2,5 0,1,2,3,4,5 0,2,3 Insufficient information 0,1,4 0,2,6

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Kay Xspre
Sep 8, 2015

Provided that all of the coefficients in the problem is the real number, the complex conjugate root theorem stated that if the polynomial has real coefficients and the complex number a + b i a+bi be the root to that polynomial, its conjugate a − b i a-bi shall also be the root , thus, the amount of complex roots will only be in the even number, which, for the polynomial of degree 5, may have only zero, two or four complex roots.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Your problem does not state that the coefficients are real. If the coefficients are complex, the number of non-real complex solutions may be 1, 3, or 5 as well.

Arjen Vreugdenhil - 5 years, 8 months ago

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