factor completely?

Factor out a 3 from each first, then factor the binomial.

3(a^2+4x-5)=3(a-1)(a+5)

and

3(a^2+3-4)=3(a-1)(a+4)

Question 1
3 (a² + 4a - 5)
3 (a + 5) (a - 1)

Question 2
3 (a² + 3a - 4)
3 (a + 4) (a - 1)

3a^2+12a-15 = (3a - 3)(a+5)

&

3a^2+9a-12 = (3a-3)(a+4)

(3a-3)(a+5)

&

3(a^2+3a-4)
3(a+4)(a-1)

1)
3a^2 + 12a - 15
= 3(a^2 + 4a - 5)
= 3(a + 5)(a - 1)

2)
3a^2 + 9a - 12
= 3(a^2 + 3a - 4)
= 3(a + 4)(a - 1)

Answer for number 1: (3a+15)(a-1) simplify you get 3(a+5)(a-1)

number 2: (3a+12)(a-1)

simplify you get...

3(a+4)(a-1)

3a^2 + 12a - 15 factors to (3a - 3)(a + 5)

3a^2 + 9a - 12 factors to (3a + 12)(a - 1)