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6 March, 14:16

How many complex zeros does the polynomial function have?

F (x) = 2x^4 + 5x^3 - x^2 + 6x-1

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  1. 6 March, 14:55
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    Two complex roots.

    Step-by-step explanation:

    F (x) = 2x^4 + 5x^3 - x^2 + 6x-1

    is a polynomial in x of degree 4.

    Hence F (x) has 4 roots. There can be 0 or 2 or 4 complex roots to this polynomial since complex roots occur in conjugate pairs.

    Use remainder theorem to find the roots of the polynomial.

    F (0) = - 1 and F (1) = 2+5-1+6-1 = 11>0

    There is a change of sign in F from 0 to 1

    Thus there is a real root between 0 and 1.

    Similarly by trial and error let us find other real root.

    F (-3) = - 1 and F (-4) = 94

    SInce there is a change of sign, from - 4 to - 3 there exists a real root between - 3 and - 4.

    Other two roots are complex roots since no other place F changes its sign
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