FInd a cubic equation with intergral coeficients that has roots 3 and 7 - i(assume the leading coeficient is 1?
Show all work and thank you soooooooo much!!! I reaaallly appreciated it!
- GeronimoLv 79 years agoFavorite Answer
x³ + bx² + cx + d = 0
Complex roots occur in pairs: 7 – i and 7 + i
then all three roots shown as product terms is:
(x – 3) • (x – 7 + i) • (x – 7 – i) = 0
... Expanding these terms ...
(x – 3) • (x² – 14x + 50) = 0 ... intermediate step
x³ – 17x² + 92x – 150 = 0
so the initial coefficient = a = 1
[b = -17] [c = 92] and [d = -150]
- Anonymous9 years ago
Ok so here is your answer.
x^3-17 x^2+92 x-150
You get this by multiplying (x-3)(x-7-i)(x-7+i)
The only tricky part to this is finding the conjugates. Its very easy, all you have to do is change the - sign in 7-i to 7+i. If you know that and how to set up factors, you just FOIL the (x-7-i)(x-7+i) first. And then do the same thing again with your answer and (x-3)Source(s): HS Algebra tutor