The number of negative roots of equation x7 + x5 - 2x4 + x3 - 3x2 + 7x - 5 = 0 is
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The number of negative roots of equation x7 + x5 - 2x4 + x3 - 3x2 + 7x - 5 = 0 is
One
Two
Three
None
Using Descartes' Rule of Signs for f(-x): (-x)^7 + (-x)^5 - 2(-x)^4 + (-x)^3 - 3(-x)^2 + 7(-x) - 5 = -x^7 - x^5 - 2x^4 - x^3 - 3x^2 - 7x - 5. There are 0 sign changes in the coefficients, so there are no negative roots.
By Descartes' Rule of Signs, we substitute x with -x to find the number of negative roots, yielding f(-x) = -x^7 - x^5 - 2x^4 - x^3 - 3x^2 - 7x - 5 = 0. Counting the sign changes in f(-x) shows that all coefficients are negative, meaning there are zero sign changes. Therefore, the equation has no negative roots.