CBSE Class 10 Answered

Using division algorithm we get

 

If (x³ + 2x² + a) is a factor of (x⁵ – x⁴ – 4x³ + 3x² + 3x + b), then remainder should be zero.
– (1 + a) x² + (3 + 3a) x + (b – 2a) = 0 = 0.x² + 0.x+0
On comparing the coefficient of x, we get
a + 1 = 0
⇒a = -1
and b – 2a = 0
⇒b =2a
b = 2(-1) = -2                                                   [va = -1]
For a = -1 and b = – 2, the zeroes ofq(x) are also the zeroes of the polynomial p(x).
q(x) = x³ + 2x² -1
p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x – 2
Divident = divisor x quotient + remainder
p(x) = (x³ +2x² -1)(x² -3x + 2)+ 0
= (x³ + 2x² -1){x² -2x – x + 2}
= (x³ + 2x² – 1) (x – 2) (x – 1)
Hence, the zeroes of p(x) are land 2 which are not the zeroes of q(x).