Question

The number of real zeroes , a polynomial of degree 5 cannot have :
1: 3
2: 5
3: 1
4: 2


Which option is correct??​

Answer 1

Answer:

Option 4th is vorrect... Imaginary roots lies in pair... If one root is imaginary then second root is also imaginary and is cojugate of first root. So only odd number of real roots are possible if degree of a polynomial is odd.

I hope this will help you.

Answer 2

The general equation of polynomial of degree 5 can be expressed as:

$ax^{5}+bx^{4}+cx^{3} +dx^{2} +ex+f$.

A polynomial of degree 5 must have exactly 5 complex zeros. If it only has real coefficients, then each nonreal zero must be paired with its complex conjugate, and so there must be an even number of nonreal zeros. Thus, there must be either 0, 2, or 4 nonreal zeros, and therefore the number of real zeros must be either 5, 3, or 1. There cannot have 2 real zeros in the conditions given.

Hence the correct option is 4: 2.

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