Question

99a^2-202ab+99b^2. factorise​

Answer 1

$99 {a}^{2} - 202ab + 99 {b}^{2}$

$99 {a}^{2} - 121ab - 81ab + 99b ^{2}$

$= 11a(9a - 11) - 9b(9a - 11b)$

$= (11a - 9a)(9a - 11b)$

The factors of

$99 {a}^{2} - 202ab + 99b ^{2}$

Are

$(11a - 9a)(9a - 11b)$

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