polynomial of degree 3 32 i SolutionAlways irrational zeros

polynomial of degree 3:

3,2 i

Always irrational zeros occur in pairs.

So, since 2-i is a zero, its additive conjugate 2+i is also a zero.

So the zeros are

3, 2-i, 2+i

So the corresponding factors are

(x-3), (x-(2-i)), (x- (2+i))

So the equation after we simplify these is,

(x-3) ((x-2) + i) ((x-2) - i)

= (x-3) ((x-2)²-i²)

= ( x-3) (x²-4x+4-(-1)) (Because i²=-1)

= (x-3) (x²-4x+5)

= x³-4x²+5x -3x²+12x-15

=x³-7x²+17x-15