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Sagot :
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.
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