Math, asked by ssjosh1967, 1 year ago

What will be its answer reply me fast??
k³+6k²+9k+4


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ssjosh1967: hmm

Answers

Answered by divyathadani
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


ssjosh1967: Thanks
ssjosh1967: Bt still i hv a doubt
ssjosh1967: What is U+ 2193.svg
divyathadani: I have another method
divyathadani: let f(k)=k^3+6k^2+9k+4
ssjosh1967: Pls do this in nb and snd me that pic
divyathadani: there is no option given for photograph
ssjosh1967: Its alright
Answered by TheIncorporealKlaus
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


ssjosh1967: Thanks alot
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