What will be its answer reply me fast??
k³+6k²+9k+4
ssjosh1967:
bs nhi mtlb nhi
Answers
Answered by
2
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)
Hope this will help u :D
[ 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)
Hope this will help u :D
Answered by
2
Let's make the polynomial a function, g
g(k) = k³+6k²+9k+4
Factors of 4 are ±1, ±2, ±4
We know that keeping a positive value will yield a large number, so let's keep k=-1
g(-1) (-1)³ + 6 × 1 + 9(-1) + 4 = 0
This means (k+1) is a factor
Now by long division, divide g(k) by (k+1)
The quotient will be : k²+5k+4
Factorizing the quotient :
k² + 4k +k + 4
= k(k+4)+1(k+4)
=(k+4)(k+1)
To write the factorized g(k), we write in the form : divisor × quotient + remainder, where remainder is zero since we're using the factors
g(k) = (k+1) (k+4) (k+1)
g(k) = (k+1)² (k+4) ... Answer
g(k) = k³+6k²+9k+4
Factors of 4 are ±1, ±2, ±4
We know that keeping a positive value will yield a large number, so let's keep k=-1
g(-1) (-1)³ + 6 × 1 + 9(-1) + 4 = 0
This means (k+1) is a factor
Now by long division, divide g(k) by (k+1)
The quotient will be : k²+5k+4
Factorizing the quotient :
k² + 4k +k + 4
= k(k+4)+1(k+4)
=(k+4)(k+1)
To write the factorized g(k), we write in the form : divisor × quotient + remainder, where remainder is zero since we're using the factors
g(k) = (k+1) (k+4) (k+1)
g(k) = (k+1)² (k+4) ... Answer
Similar questions