Factories:-
$6 n + 12 n^{2} + 21 n$

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Solution

We have,
$6 n + 12 n^{2} + 21 n$
$= 12 n^{2} + 27 n$

On factorization we have,
$= 3 n (4 n + 9)$

Hence, this is the answer.

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Similar questions

Q. Factories:-

6 n + 12 n^{2}

Find the sum of $1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . . . . n {.1}^{2}$

The least positive integer k for which the value $k \times n^{2} (n^{2} - 1^{2}) (n^{2} - 2^{2}) . . . . (n^{2} - (n - 1)^{2})$ turns into a factorial of some positive integer is

Q. Factorise:

(i) 7m + 21n

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(9) 4m² + 19mn + 12n²

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Factories:-6n+12n2+21n

We have,6n+12n2+21n=12n2+27nOn factorization we have,=3n(4n+9)Hence, this is the answer.

Factories:-
$6 n + 12 n^{2} + 21 n$

We have,
$6 n + 12 n^{2} + 21 n$
$= 12 n^{2} + 27 n$

On factorization we have,
$= 3 n (4 n + 9)$

Hence, this is the answer.