Question

factorise using suitable identity 9a square - 4b square - 6a + 1

Answer 1

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${9a}^{2} - {4b}^{2} - 6a + 1$

Rearranging the terms

$({9a}^{2} - 6a + 1) + ( - 4b {}^{2} ) \\ \\ ( {9a}^{2} - 6a + 1) - {(4b)}^{2} \\ \\ (( {3a)}^{2} - 2 \times 3a \times 1+ {(1)}^{2} ) - {(2b)}^{2}$
Now the terms in the first brackets are in format (a - b)^2 .

$(3a - 1) {}^{2} - {(2b)}^{2}$
Now the two terms in separate brackets are in format a^2 - b^2 . We will factorise in the format (a+b) (a-b).

$((3a - 1) + (2b)) \: \: ((3a - 1) - (2b) \\ \\ (3a - 1 + 2b) \: \: (3a - 1 - 2b)$

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#AADI 93

Answer 2

a square + 6a+ 9 answer