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).
a^(3) + a^(2) Factors to a^(2) [ a^(2) + 1 ] NB THe terms in the square brackets do NOT factor. Reason, Any ywo squared terms with a positive (+) between then does NOT factor Think of a^(2) + 1 as a^(2) + 1^(2) . A pair of squared terms with a negative between them DOES factor. So if you initial expression was a^(4) - a^(2) Then a^(2) [ a^(2) - 1^(2)] a^(2)[a - 1)(a + 1) Fully factored. Note the two different signs.
-18
a2-b2 = (a-b)(a+b)
a2 - 4a + 4
-18
(a2+2b2-2ab)(a2+2b2+2ab)
a2-5a-14 = (a-7)(a+2) when factored
(x+a) (x-a)
(a + f)(a - f)
a2 + 12a + 27 = (a + 3)(a + 9)
a2+9a+20 = (a+4)(a+5) when factored
16 - a2