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Question

Find the HCF of p(x)=x2x12 and q(x)=x29.

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Solution

We have
p(x)=x2x12=(x4)(x+3), q(x)=x29=(x3)(x+3)
Here, the common factor is x+3 and the lowest power of x+3 in p(x) and q(x) is x+3 itself. Hence x+3 is the HCF of p(x) and q(x).

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