Solve:
c3 + c2 - 7c - 3 = 0, given root -3
2006-12-31 04:53:55 · 3 answers · asked by lazarus d 1 in Science & Mathematics ➔ Mathematics
Synthetic division:
-3 | +01 +01 -07 -03
---| +00 -03 +06 +03
-------------------------
---| +01 -02 -01 +00
So after dividing by (x + 3) you're left with:
c² - 2c - 1
Use the quadratic:
c = (2 ± √(4 - 4(-1)))/2
= (2 ± √(4 + 4))/2
= (2 ± √8)/2
(2 ± 2√2)/2
= 1 ± √2
So the solutions are -3, 1 + √2, 1 - √2.
2006-12-31 04:58:28 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
dividing by c+3 synthetically
1 1 -7 -3
0 -3 6 3
1 -2 -1
so the quotient=c^2-2c-1
c=[2+/-rt(8)]/2=1+/-rt2
so the roots are -3,1+rt2,1-rt2
2006-12-31 05:21:38 · answer #2 · answered by raj 7 · 0⤊ 0⤋
c3 + c2 - 7c - 3 = 0
(c+3)(c2-2c-1)=0
c=-3 , c=-0.414, c=2.414
2006-12-31 05:03:15 · answer #3 · answered by Mena M 3 · 0⤊ 0⤋