How can you check if 3+i is a zero of the equation x^4-6x^3+6x^2+24x-40?

Answer 1

One way is to substitute x=3+i into the equation and evaluate:
x^4-6x^3+6x^2+24x-40.....(1)
=(3+i)^4-6(3+i)^3+6(3+i)²+24(3+i)-40
=28+96i - 6(18+26i) + 6(8+6i) + 24(3+i) -40
=96i-156i+36i+24i + 28-108+48+72-40
=0i+0
=0

Therefore (3+i) is a zero of the given equation.
Note that the complex conjugate of the given zero is also a zero if the original equation has real coefficients. So as a bonus, 3-i is also a zero of (1).

Another way is to factorize the given expression on the left-hand side, namely
(x-2)*(x+2)*(x^2-6*x+10)
Use the quadratic formula to find the two complex roots in the last term of the factorization to give 3±i as the remaining zeroes.