For which values of a and b, the zeroes of q (x) = x^{3} + 2x^{2} + a are also the zeros of the polynomial p(x) = x^{5} - x^{4} - 4x^{3} + 3x + b? Which zeroes of p(x) are not the zeroes of p (x)?

Let P(x) = x^{5} - x^{4} - 4x^{3} + 3x + b and q (x) = x^{3} + 2x^{2} + a

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

r(x) = - (a + 1)x^{2} + 3(1 - a)x + b - 2a = 0 and g(x) = x^{2} - 3x + 2

By comparing the coefficients

- (a + 1) = 0 , (1 - a) = 0

⇒ a = - 1

Also, b - 2a = 0

= b = 2a

= 2(- 1) = - 2

⇒ b = - 2

q (x) = x^{3} + 2x^{2} - 1

x^{3} + 2x^{2} - 1 = 0

x^{3} + x^{2} + x^{2} - 1 = 0

x^{2}(x + 1) + (x^{2} - 1) = 0

x^{2}(x + 1) + (x + 1)(x - 1) = 0

(x + 1)(x^{2} + x - 1) = 0

g(x) = x^{2} - 3x + 2

x^{2} - 3x + 2 = 0

x^{2} - (x + 2x) + 2 = 0

x(x - 1) - 2(x - 1) = 0

(x - 1)(x - 2) = 0

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

P(x) = x^{5} - x^{4} - 4x^{3} + 3x - 2 = q(x).g(x) + 0

= (x^{3} + 2x^{2} - 1) (x2 - 3x + 2)

= (x + 1) (x2 + x - 1) (x - 1) (x - 2)

∴ 1 and 2 are the zeroes of p(x) that are not in q(x).

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