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The expression is faulty because we don't have + or - before 10a. Also have to assume a2 is a2 . Use the standard formula for solving a quadratic = 0, to get two answers p and q. Then the factorization is (a-p)(a-q).
a2-b2 = (a-b)(a+b)
(a2+2b2-2ab)(a2+2b2+2ab)
16 - a2
5(a2-3)
ANS:(a-3)(a+1)
( a2 ) ( a2+1 )
-18
a2 - 4a + 4
(x+a) (x-a)
(a + f)(a - f)
a2-5a-14 = (a-7)(a+2) when factored
a2+9a+20 = (a+4)(a+5) when factored
a2 + 12a + 27 = (a + 3)(a + 9)