Show that the polynomial x^8 -x^7+x^2 -x+15=0 Has no real roots.

Need this for an exam Urgently any kind of help would be appreciated

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## Comments

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TopNewestLet K = x^8-x^7 + x^2 - x = x(x-1)(x^6+1)

Observe that x must be in (0,1) for K to be negative

x^6+1 takes values in (1,2) for x in the range (0,1)

The minimum value of (x)(x-1) is -1/4

Thus K >= -1/2

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From the above, K + 15 >= 29/2

Thus x^8 - x^7 + x^2 - x + 15 is always positive and thus has no real roots.

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descartes rule of signs only must a limit on the maximum real roots.i dont think thats the only idea to be applied here.

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I do not know about how to prove it has no real roots. You can show that it has no rational roots.

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actually if we are given with a polynomial where f(x) = 0 ., then for f(x) to have no real roots, the condition is that f(-x) should not have ny change of signs...... so in the given polynomial f(-x) = (-x)^8 -(-x)^7 +(-x)^2 -(-x)+15 => f(-x) = x^8 +x^7+x^2+x+15

so thr is no change of signs in f(-x)....and hence there r no real roots , all of the roots r imaginary...

here i've used the descartes rule of signs

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there can be 4 or 2 and maybe 0 positive real roots for the polynomial according to descarte's theorem.

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But that is only concluded for negative roots, what about real positive roots?

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no thts not correct actually......the maximum number of negative real roots of a polynomial equation f(x)=0 is the number of change of signs from positive to negative and negative to positive in f(-x)

and the number of positive real roots of a polynomial equation f(x)=0 is the number of change of signs from positive to negative and negative to positive in f(x)

and if f(-x) has got no sign change, then the polynomial equation has complex roots..

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so wht can u conclude from this?.... i guess thr is a flaw in the question

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so u've finished ur 12th this yr?...r u a cbse student?

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Comment deleted Jul 14, 2013

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Decartes' rule of signs? Maybe you could use that.

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