For which values of a and b, the zeroes of q (x) = x3 + 2x2 + a are also the zeros of the polynomial p(x) = x5 - x4 - 4x3 + 3x + b? Which zeroes of p(x) are not the zeroes of p (x)?
Let P(x) = x5 - x4 - 4x3 + 3x + b and q (x) = x3 + 2x2 + a
Dividend = (divisor) (quotient) + remainder
⇒ p(x) = g(x).q(x) + r(x)
r(x) = - (a + 1)x2 + 3(1 - a)x + b - 2a = 0 and g(x) = x2 - 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) = x3 + 2x2 - 1
x3 + 2x2 - 1 = 0
x3 + x2 + x2 - 1 = 0
x2(x + 1) + (x2 - 1) = 0
x2(x + 1) + (x + 1)(x - 1) = 0
(x + 1)(x2 + x - 1) = 0
g(x) = x2 - 3x + 2
x2 - 3x + 2 = 0
x2 - (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) = x5 - x4 - 4x3 + 3x - 2 = q(x).g(x) + 0
= (x3 + 2x2 - 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).