The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is ![6 a b^{2}(a+2)(2 a-1)](https://tex.z-dn.net/?f=6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29)
<u>Solution:</u>
Given, expression is ![12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}](https://tex.z-dn.net/?f=12%20a%5E%7B3%7D%20b%5E%7B2%7D%2B18%20a%5E%7B2%7D%20b%5E%7B2%7D-12%20a%20b%5E%7B2%7D)
We have to factorize the given expression completely.
Now, take the expression
![12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}](https://tex.z-dn.net/?f=12%20a%5E%7B3%7D%20b%5E%7B2%7D%2B18%20a%5E%7B2%7D%20b%5E%7B2%7D-12%20a%20b%5E%7B2%7D)
Taking
as common term,
![b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%2812%20a%5E%7B3%7D%2B18%20a%5E%7B2%7D-12%20a%5Cright%29)
Taking "a" as common term,
![b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%2812%20a%5E%7B2%7D%2B18%20a-12%5Cright%29%5Cright%29)
Taking "6" as common term,
![b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%286%5Cleft%282%20a%5E%7B2%7D%2B3%20a-2%5Cright%29%5Cright%29%5Cright%29)
Splitting "3a" as "4a - a" we get,
![b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%286%5Cleft%282%20a%5E%7B2%7D%2B4%20a-a-2%5Cright%29%5Cright%29%5Cright%29)
![\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7Bb%5E%7B2%7D%28a%286%282%20a%28a%2B2%29-1%28a%2B2%29%29%29%29%7D%20%5C%5C%5C%5C%20%7Bb%5E%7B2%7D%28a%286%28%28a%2B2%29%20%5Ctimes%282%20a-1%29%29%29%29%7D%20%5C%5C%5C%5C%20%7B6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29%7D%5Cend%7Barray%7D)
Hence, the factored form of given expression is ![6 a b^{2}(a+2)(2 a-1)](https://tex.z-dn.net/?f=6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29)