For which values of a and b are the zeroes of q(x) = (x³ + 2x² + a) also the zeroes of the polynomial p(x) = (x⁵ – x⁴ – 4x³ + 3x² + 3x + b)? Which zeroes of p(x) are not the zeroes of q(x)?

q(x) = x³ + 2x² + a

p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + b

for zeroes of q(x) to also be zeroes of p(x), p(x) should be divisible by q(x)

dividing p(x) by q(x)

As q(x) divides p(x) the remainder should be zero

⇒ –(a + 1)x² + (3a + 3)x + b – 2a = 0 …(1)

Note: here we do not have to Find the zeroes of polynomial (1) because x is the variable for p(x) and q(x) and not (1) and x won’t be 0(why? can check by substituting x = 0 in p(x) and q(x)) hence the coefficients and constant part of (1) has to be 0

Equating coefficient of x² to 0

⇒ –(a + 1) = 0

⇒ a = –1

Equating constant to 0

⇒ b – 2a = 0

⇒ b = 2a

Put a = 1

⇒ b = 2

Hence p(x) and q(x) becomes

q(x) = x³ + 2x² + 1

p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + 2

degree of p(x) is 5 hence it will have 5 zeroes

using dividend = divisor × quotient + remainder

⇒ p(x) = q(x) × (x² – 3x + 2)

Remaining zeroes of p(x) which are not same for p(x) and q(x) can be found when x² – 3x + 2 = 0

⇒ x² – 3x + 2 = 0

⇒ x² – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 1)(x – 2) = 0

⇒ x – 1 = 0 and x – 2 = 0

⇒ x = 1 and x = 2

Hence value of a and b for p(x) to have zeroes of q(x) are 1 and 2 respectively and the zeroes of p(x) which are not zeroes of q(x) are x = 1 and x = 2