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¯¯¯

- k3 .+ k2_ U+2193.svg U+2193.svg

5k2 + 9k U+2193.svg

- 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)

U+2196.svg(from)

∴ the factors are : (k+1) (k+1)(k+4)

(from)U+2193.svg k2+5k+4

(k+1) √¯¯k3+ 6k2+ 9k+ 4¯¯¯

∴ Ans : (k+1)(k+1)(k+4)