Question

xcube-bxsquare+2b+5/x-bx-b​

Answer 1

Answer:

Let p(x) = x3 + ax2 + bx +6

 (x-2) is a factor of the polynomial x3 + ax2 + b x +6

p(2) = 0                                                                                       

p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0

7 +2 a +b = 0

b = - 7 -2a -(i)                                                                              

x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.

p(3) = 3                                                                                      

p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3

11+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.