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Question

Find the HCF of p(a)=2a27a+3 and q(a)=2a2+5a3.

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Solution

Here
p(a)=2a27a+3=2a26a+3=2a(a3)1(a3)=(a3)(2a1)
and
q(a)=2a2+5a3=2a2+6aa3=2a(a+3)1(a+3)=(a+3)(2a1)
From this we conclude that the HCF of p(a) and q(a) is 2a1.

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