2006-10-24 16:03:31 · 7 answers · asked by Trinity 1 in Science & Mathematics ➔ Mathematics
x^3-x^2+16x-16 = x^2(x - 1) + 16(x - 1)
= (x - 1)(x^2 + 16)
[= (x - 1)(x - 4i)(x + 4i) if complex numbers are OK. If not stop above)]
Roots are 1 (and ± 4i if complex roots are also being sought)
2006-10-24 16:15:09 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
I'm only doing this because I like math problems.
x^3-x^2+16x-16=(x-1)(x^2+16)
Your roots are x=1 and x^2=-16. Since a square can not be negative, if x^2=16, then x=4i.
So, your roots are 1 and 4i. Hope that helps.
2006-10-24 23:20:34 · answer #2 · answered by hotstepper2100 3 · 0⤊ 0⤋
x^3-x^2+16x-16
=(x-1)(x^2+4^2)
this means the only root is x=1 since x^2+4^2 cannot be factored.
2006-10-24 23:13:15 · answer #3 · answered by abhiiii 2 · 0⤊ 0⤋
x^3 - x^2 + 16x - 16 = 0
= x^2(x - 1) + 16(x - 1) = 0
= (x - 1)(x^2 + 16) = 0
= (x -1) = 0 and (x^2 + 16) = 0
= x = 1 and x^2 = -16 or x = sqr(-16) orx = 4i or -4i
2006-10-25 00:28:00 · answer #4 · answered by Subhash G 2 · 0⤊ 0⤋
x³ - x² + 16x - 16 = 0
x²(x-1) + 16 (x-1) = 0
(x-1) (x² + 16) = 0
x - 1 = 0
x = 1
or x² = -16
x = ± 4i
where i = â -1
then
x Ð { 1 , 4i , -4i }
2006-10-25 00:11:35 · answer #5 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
by syntethic devision
1 -1 16 -16 \_1
____1___0___16_____
1 0 16 0
one root is 1
the remaining is x+16=0
means imaginary roots of +/-16i
2006-10-24 23:10:23 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Do your own homework.
2006-10-24 23:10:36 · answer #7 · answered by qtrn2005 3 · 0⤊ 0⤋