Question

IF $x^3+a{x}^2+bx+6$ has $x-2$ as a factor and leaves a remainder 3 when divided by x-3, fidn the value of a and b.

Answer 1

Let f(x) = $x^3+a{x}^2+bx+6$

By factor theorem we get f(2) = 0 [since x-2 = 0 then x= 2]

=> $(2{)}^3+a(2{)}^2+b\left(2\right)+6=0$

=>8 + 4a + 2b + 6 = 0

=> 4a + 2b= -14

=> 2(2a + b) = -14

=>2a + b = -7 .............equation (1)

In the second case by remainder theorem we get, f(3) = 3 [since x-3= 0 then x= 3]

=>$(3{)}^3+a(3{)}^2+b\left(3\right)+6=3$

=> 27 + 9a +3b + 6 = 3

=> 33 + 9a + 3b = 3

=> 9a + 3b = -30

=> 3(3a + b) = -30

=>3a + b = -10 .............equation (2)

b= -7 -2a

Substituting the value of b in equation (2) we get,

3a -7 -2a= -10

=> a-7 = -10

=> a = -3

Substituting the value of a in equation (1) we get, 2(-3) +b = -7

=> -6 + b = -7

=>b = -1

Therefore the respective values of a and b are -3 and -1.