Let P(x)=x−3+4i3−4ix−7i5+6i−x7−2i−7−2i .The number of values of x for which P(x) = 0 is
0
1
2
3
Using R2→R2−R1,R3→R3+R1,
We get
P(x)=x−3+4i3−4i03−11i2+10i04+2i−4−6i
As P(x) is a linear polynomial, P(x) = 0 for exactly one value of x.