f(x)-7xpower of 5+6xpower of 4 +2xpower of 3 +9xpower of 2+x+5Solution.docx

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f(x)=7xpower of 5+6xpower of 4 +2xpower of 3 +9xpower of 2+x+5 Solution 7x^5+6x^4+2x^3+9x^2+5 = 0 From decart\'s sign rule there is no sign change so no +ve real roots replacing x by -x we get f(-x)=- 7x^5+6x^4-2x^3+9x^2+5 = 0 so there is only 3 sign change so maximum possible -ve roots is 3 so remaining 2 are complex root. finally complex root will occur in conjugate so 4 complex roots in total..so remaining 5-4=1 will be the actual number of -ve root. .

f(x)=7xpower of 5+6xpower of 4 +2xpower of 3 +9xpower of 2+x+5
Solution
7x^5+6x^4+2x^3+9x^2+5 = 0
From decart's sign rule there is no sign change so no +ve real roots
replacing x by -x we get
f(-x)=- 7x^5+6x^4-2x^3+9x^2+5 = 0
so there is only 3 sign change so maximum possible -ve roots is 3
so remaining 2 are complex root. finally complex root will occur in conjugate so 4 complex roots
in total..so remaining 5-4=1 will be the actual number of -ve root.

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