First you can factor a^2 from the first two terms and 4 from the last two terms.
(a^2)(a-2)+4(a-2)
You can see that they both have (a-2) as a factor.
So it is (a^2+4)(a-2)
a3 - 2a2 + 4a - 8 = a2(a - 2) + 4(a - 2) = (a - 2)(a2 + 4)
the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)
kutta
a3+b3
There are no whole solutions
a3 - 2a2 + 4a - 8 = a2(a - 2) + 4(a - 2) = (a - 2)(a2 + 4)
(a - 2)(a^2 + 4)
(a + 4)(a^2 - 4a + 16)
4a4 + 11a3 - 47a3 - 4a + 5 =4a4 - 36a3 - 4a + 5 =4a (a3 - 9a2 - 1) + 5
a3-4a = a(a2-4) when factored
2a3 - 128 = 2*(a3 - 64) = 2*(a -4)*(a2 + 4a + 16)
a(a^2 - 9a + 3)
A3+b3
(a - 1)(a^2 + a + 1)
the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)
kutta
Just choose the smaller exponent. The GCF of a3 and a5 is a3