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Alenkinab [10]
3 years ago
12

Is the HCF of a^3-8, a^4+16 and a^3+2a^2+4a just 1? Please answer........

Mathematics
1 answer:
Flauer [41]3 years ago
3 0
First we find the factors
a^3-8 is the differnce between 2 perfect cubes and is factored out to
(a-2)(a^2+2a+4)
the factors are (a-2) and (a^2+2a+4)

a^4+16
this is not factorable, unless you mistiped it and it was really a^4-16
the factor is 1


a^3+2a^2+4a
factor out the a
a(a^2+2a+4)
this is the most factored out form
the factors are (a) and (a^2+2a+4)


there are no common factors except 1 so 1 is the answer
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