Find a third-degree polynomial equation with rational coefficients that has roots –4 and 6 + i. (1 point)x3 – 8x2 – 11x + 148 = 0
x3 – 8x2 – 12x + 37 = 0
x3 – 12x2 + 37x = 0
x3 – 8x2 – 11x = 0

Question

Find a third-degree polynomial equation with rational coefficients that has roots –4 and 6 + i.  (1 point)x3 – 8x2 – 11x + 148 = 0
x3 – 8x2 – 12x + 37 = 0
x3 – 12x2 + 37x = 0
x3 – 8x2 – 11x = 0

tkhunny · Answer

It has REAL coefficients, so any Complex Roots must be in Conjugate pairs.

-4 ==> (x+4)
6+i ==> ALSO 6-i ==> (x-(6+i))(x-(6-i))

A little algebra remains.

anonymous · Answer

so it is a

anonymous · Answer

@tkhunny

tkhunny · Answer

You're going to have to tell me.  How did you get that?