how to split a cubic polynomial like k3+6k2+9k+4??
By using trial and error method,
Let f(k) = k3+ 6k2+ 9k+4
[ find a 'k' that makes the equation = 0 ]
Substitute k = 1,
We get, 1+6+9+4 ≠ 0
Substitute k = -1,
We get -1+6 - 9+4 = 0
[NOTE: Substitute k = 1, -1, 2, -2, etc until we get the above eqn. = 0 .
Here if we take k = -1 we get the above eqn. = 0 . So we take (k+1) ]
k2+5k+4
(k+1) √¯¯k3+ 6k2+ 9k+ 4¯¯¯
- 5k2 + 5 k __
4k + 4
- _ 4k + 4_
0
Now, factorising the quotient : k2+5k+4
k2+5k+4 Sum = 5 , Product = 4 ∴ (4,1)
= k2 +4k + k + 4
= k(k+4) + 1(k+4)
= (k+1)(k+4)
∴ the factors are : (k+1) (k+1)(k+4)
(k+1) √¯¯k3+ 6k2+ 9k+ 4¯¯¯
∴ Ans : (k+1)(k+1)(k+4)
Hope this will help u :D