How do you factor #81c²+198c+121#?

1 Answer
EZ as pi
Jul 3, 2016

#(9x + 11)(9x +11)#

Explanation:

There is no common factor - this can be seen because the only factors of 81 are multiples of 3 and the only prime factor of 121 is 11.

We need to find factors of 81 and 121 which ADD to give 198.
Exploring possible combinations gives 9 x 11 and 9 x 11 = 99 + 99 = 198.
Cross multiplying and adding gives the required total of 198

#9" "11 rArr 9 xx 11 = 99#
#9" "11 rArr 9 xx 11 = 99# Sum is 198.

The signs in the bracket will be the same, they are both positive,

#(9x + 11)(9x +11)#

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