Given that the zeroes of q(x)=x3+2x2+a are also the zeroes of the polynomial p(x)=x5−x4−4x3+3x2+3x+b i.e., q(x) is a factor of p(x). then, we use a division algorithm
⇒x3+2x2+ax2−3x+2√x5−x4−4x3+3x2+3x+b
⇒x5+2x4+ax2−3x4−4x3+(3−a)x2+3x+b
⇒−3x4−6x3−3ax2x3+(3−a)x2+(3+3a)x+b
⇒2x3+4x2+2a−(1+a)x2+(3+3a)x+(b−2a)
if (x3+2x2+a) is a factor of (x5−x4−4x3+3x2+3x+b), then remainder should be zero.........(Factor theorem)
i.e., −(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
For a = -1 and b = -2, the zeroes of q(x) are also the zeroes of the polynomial p(x)
∴ 1(x)=x3+2x2–1
and p(x)=x5−5x4−4x3+3x2+3x−2
now, dividend = divisor × quotient + remainder
p(x)=9x3+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 1 and 2 which are not the zeroes of q (x).