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

i think its a am i right

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

i think its a am i right

x3 – 8x2 – 11x + 148 = 0
x3 – 8x2 – 12x + 37 = 0
x3 – 12x2 + 37x = 0
x3 – 8x2 – 11x = 0

campbell_st · Answer

use the factor theorem 

and find f(-4) for each... if f(-4) = 0 then x = -4 is a root

then when you have a polynomial do some synthetic division or polynomial  division (x + 4) to see if the quadratic has a root 6 + i

anonymous · Answer

ya i think its a am i right @campbell_st

campbell_st · Answer

which one had f(-4) = 0..?

campbell_st · Answer

oops yes it is A... so I'd say thats it..