It cannot factor without imaginary numbers
a2 - 4a + 4
(a + b)(b - 2c)
the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)
as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]
4(16a4 + b2)
(a + b)(b + c)
It cannot factor without imaginary numbers
a2 - 4a + 4
(a+b+c) (a+b-c)
a(b+3)+b(b+3)
(a + b)(b - 2c)
(a + b)(b - 2c)
the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)
(b + 2)(b + 2) or (b + 2)2
as a2+b2=(a+ib)(a-ib), b2+2=[b - i sqrt(2)] [b + i sqrt(2)]
(-8 + b2) - (5 + b2) = -8 + b2 - 5 - b2 = -13