A polynomial P ( x ) has degree 3 and leading coefficient 7 . Two of its roots are − 2 i and 2 .

Question

ind a possible formula for . You do not need to simplify/distribute your answer.

Answer 1

3 roots are 2i, -2i, and 2

the imaginary roots come in conjugate pairs

to find the factors change the sign of the roots and stick an x in front of them.

(x-2)

(x-2i)

(x+2i)

multiply them all together

(x-2i)(x+2i) = x^2 +4

also multiply by 7 to get a leading coefficient of 7

(x^2+4)(x-2)(7) = 7(x^3-2x^2 +4x - 8)

P(x) = 7x^3-14x^2 + 28x -56

P(x) = 7(x-2)(x-2i)(x+2i)

set the factors = 0 to solve for the roots or zeroes

x-2 = 0

x =2

x-2i=0

x = 2i

x+2i=0

x =-2i