Question

Find the value of a and b so that the polynomials p(x) and g(x)
have (x + 1)(x - 2) as their HCF. p(x) = (x2 + 3x + 2)(x2 + x + a)
q(x) = (x2 - 3x + 2)(x2 – 3x + b).​

Answer 1

Answer:

(x+1)(x−4) is a factor of both p(x) and q(x)

p(x)=(x2+3x+2)(x2−7x+a)

or p(x)=(x+1)(x+2)(x2−7x+a)

We know that (x+1)(x−4) is a factor of p(x)

∴p(4)=0 and p(−1)=0

∴5⋅6⋅(16−28+a)=0

∴a=12

q(x)=(x2−x−12)(x2+5x+b)

or q(x)=(x−4)(x+3)(x2+5x+b)

We know that (x+1)(x−4) is a factor of q(x)

∴q(−1)=0 and q(4)=0

∴−5⋅2⋅(1−5+b)=0

∴b=4

Answer: a=12 and b=4

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