How many complex roots does the polynomial equation below have?

X^7-3x^5+4^2-1=0

4

Step-by-step explanation:  5 22 8 18 e

Graph the function and look for sign changes in the slope  (easiest)

  of the 7 roots 3 are real and 4 are complex

or the perferred way is to apply Descartes rule of signs

x^7-3x^5+4x^2-1=0        a typo in your function  attached photos shows 4x²

(x^7) - (3x^5) +(4x^2) - (1)     sign changes I see 3

then to find a bound on the number of negative roots

(-x^7) - (-3x^5) +(4x^2) - (1)    = -x^7 + 3x^5 + 4x^2 - 1     I see 2 sign changes ??

there is at least 1 real root (due the odd highest power)

and either 0, 2, 4 or 6 complex roots

so 5 and 7 are out

still confused I used a polynomials root calculator

it said ther are three real roots  (confirmed by the graph of the function)

  and four comple roots (which I can't figure out from the graph)

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

idk

step-by-step explanation: