Factor Theorem-11

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Find the values of a and b so that the polynomials P(x) and Q(x) have (x + 1) (x + 3) as their HCF,
where P(x) = (x2 + 3x + 2)(x2 + 2x + a), and Q(x) = (x2 + 7x + 12)(x2 + 7x + b).

Solution: P(x) = (x2 + 3x + 2)(x2 + 2x + a) = (x + 1)(x + 2)(x2 + 2x + a)

(Factorizing the trinomial x2 + 3x + 2)

Similarly, Q(x) = (x2 + 7x + 12)(x2 + 7x + b) = (x + 3)(x + 4)(x2 + 7x + b)

HCF = (x + 1)(x + 3)

=> (x + 1) and (x + 3) are factors of P(x) and Q(x).

Therefore, (x + 3) is a factor of P(x). Therefore, P( -3) = 0

=> (-3 + 1)(-3 + 2)(9 - 6 + a) = 0

=> ( -2)( -1)(3 + a) = 0

=> 2(3 + a) = 0; => a = -3

Now since (x + 1) is a factor of Q(x), therefore, Q( -1) = 0

=> ( -1 + 3) ( -1 + 4) (1 -7 + b) = 0

=> (2) (3) (b - 6) = 0

=> b = 6

Hence, a = -3, b = 6

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