An important question from examination point of as it has been already asked in previous year paper of 2005 in which we are asked to find the values of a and b if (x – 2) is a factor of the expression x3+ax2+bx+6. when this expression is divided by (x – 3) , it leaves the remainder 3

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Let p(x) = x

^{3}+ ax^{2}+ bx +6(x-2) is a factor of the polynomial x

^{3}+ ax^{2}+ b x +6p(2) = 0

p(2) = 2

^{3}+ a.2^{2}+ b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 07 +2 a +b = 0

b = – 7 -2a -(i)

x

^{3}+ ax^{2}+ bx +6 when divided by (x-3) leaves remainder 3.p(3)=3

p(3) = 3

^{3}+ a.3^{2}+ b.3 +6= 27+9a +3b +6 =33+9a+3b = 311+3a +b =1

=> 3a+b =-10

=> b= -10-3a -(ii)

Equating the value of b from (ii) and (i) , we have

(- 7 -2a) = (-10 – 3a)

a = -3

Substituting a = -3 in (i), we get

b = – 7 -2(-3)

= -7 + 6

= -1

Thus the values of a and b are -3 and -1 respectively.