Question

for the equation, find the number of complex roots, the possible number of real roots and the possible rational roots. x^7-2x^6+3x^2-2x+5=0

Answer 1

lol why am i keep seeing questions like these

Answer 2

I WAS JUST GONNA SAY THAT!

Answer 3

lol

Answer 4

so is there an answer or what

Answer 5

lol sorry

Answer 6

im lost

Answer 7

where?

Answer 8

lol

Answer 9

The only possible rational roots are \(\pm 5\) (by the rational roots theorem); I'll leave it to you to decide if they are actually roots of the polynomial.

As for the possible number of real/complex solutions.....

By Descarte's Rule of Signs, the polynomial \(f(x) = x^7-2x^6+3x^2-2x+5\) can have 4, 2, or 0 positive real solutions (you determine this by looking at the number of sign changes between the terms in the polynomial; there are 4 sign changes, so the rule says we can have 4, 2, or 0 positive solutions). We then note that \(f(-x) = -x^7-2x^3+3x^2+2x+5\) and thus by Descarte's Rule of Signs, \(f(x)\) has exactly 1 negative real root (since there's only one sign change in the coefficients). Hence, DRS guarantees that we will have either 1+4 = 5, 1+2 = 3, or 1+0= 1 real solutions.

Furthermore, by the fundamental theorem of algebra, \(f(x)\) has exactly 7 roots (counting both real and complex roots). Coupling this result with the fact that Descarte's rule of signs guarantees either 1,3,or 5 real solutions for our polynomial equation, we now see that the polynomial can have either 5 real solutions & 2 complex solutions, 3 real solutions & 4 complex solutions, or 1 real solution and 6 complex solutions (since complex solutions come in conjugate pairs).

Does this make sense? :-)