d. 3
(a - 6)(a + 3)
a3 + b3 = (a + b)(a2 - ab + b2) a3 - b3 = (a - b)(a2 + ab + b2)
a2 + 28a + 27 = 0 (a + 1)(a + 27) = 0 a = -1 or -27 However, the second line is the factorised form, the third is what a is equal to
(a + 9)(a - 9)
(a + 4)(a - 5)
( a2 ) ( a2+1 )
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)
-18
-18
a2 - 4a + 4
(x+a) (x-a)
(a + f)(a - f)
(a2+2b2-2ab)(a2+2b2+2ab)
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)