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Anonymous
(b + 8)(b + 4)
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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)]