asked 5.1k views
3 votes
Is the HCF of a^3-8, a^4+16 and a^3+2a^2+4a just 1?
Please answer........

1 Answer

3 votes
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
answered
User CFIFok
by
8.2k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.