Find the number of positive real roots for equation $3x^5 - 4x^4 - 42x^3 + 56x^2 + 27x - 36 = 0$. If all the roots of above given equation are real.

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Solution

Solution: Let $f (x) = 3 x^{5} - 4 x^{4} - 42 x^{3} + 56 x^{2} + 27 x - 36$

According to Descartes' rule of signs,

Total number of positive real roots = No. of charges in the sign occurs at any particular terms.

$f (x) = 3 x^{5} - 4 x^{4} - 42 x^{3} + 56 x^{2} + 27 x - 36$

$+$ $-$ $-$ $+$ $+$ $-$

Change Change Change

No. of positive real roots $= 3$ .

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Find the number of positive real roots for equation $3x^5 - 4x^4 - 42x^3 + 56x^2 + 27x - 36 = 0$. If all the roots of above given equation are real. __

Solution: Let f(x)=3x5−4x4−42x3+56x2+27x−36 According to Descartes' rule of signs, Total number of positive real roots = No. of charges in the sign occurs at any particular terms. f(x)=3x5−4x4−42x3+56x2+27x−36 + − − + + − Change Change Change No. of positive real roots =3.

Find the number of positive real roots for equation $3x^5 - 4x^4 - 42x^3 + 56x^2 + 27x - 36 = 0$. If all the roots of above given equation are real.

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