Answer: b) 4
Step-by-step explanation:
[tex]x^7-3x^5+4x^2-1=0\\\\\text{The only possible rational roots are 1 and -1.}\\\text{Use synthetic division to see if either of these have a remainder of 0.}\\\text{Remember to use 0's for the missing exponents.}\\\\1\ |\ 1\qquad 0\quad -3\qquad 0\qquad 0\qquad 4\qquad 0 \quad -1\\.\ \ \underline{| \downarrow \qquad 1\qquad 1\quad -2\quad -2\quad -2\ \quad 2\qquad 2}\\.\quad 1\ \qquad 1\quad -2\quad -2\quad -2\qquad 2\ \quad 2\qquad \boxed{\ 1\ }\leftarrow \text{Remainder }\neq0[/tex]
[tex]-1\ |\ 1\qquad 0\quad -3\qquad 0\qquad 0\qquad 4\qquad 0 \quad -1\\.\ \quad \underline{| \downarrow \quad -1\qquad 1\qquad 2\quad -2\qquad 2\ \quad -6\quad \ 6}\\.\qquad 1\quad -1\quad -2\ \quad 2\quad -2\qquad 6\ \quad -6\quad \boxed{\ 5\ }\leftarrow \text{Remainder }\neq0[/tex]
Since there are no rational roots and there cannot be an odd number of complex roots, the next step is to graph it (see below). There are 3 x-intercepts and the polynomial has degree 7 so there are 7 - 3 = 4 complex roots
