Given a2 + 11b + 11ab + a =(a × a + a × 1 + (11b) × 1 + (11b) × a)
Taking a common from 1st and 2nd term and (11b) from 3rd and 4th,
⇒ a2 + 11b + 11ab + a = a (a + 1) + (11b)(1 + a)
⇒ a2 + 11b + 11ab + a = (a + 1)(a + 11b)
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