Question

if(2x^3+ax^2+bx-6)has (x-2) as a factor and leaves 36 as a reminder when divided by (x-3) find the value of a and b

Answer 1

Let x - 2 = 0
 or x = 2
Using remainder theorem
p(2) = 2(2)³ + a(2)² +b(2) - 6
       = 16 + 4a + 2b - 6
       = 10 + 4a +2b 
as it is a factor then
   10 + 4a +2b = 0
or 4a = - 10 - 2b
or a = -10 - 2b /4 ----- (i)

Let x - 3 = 0
or x = 3
Using remainder theorem
p(3) = 2(3)³ + a(3)² + b(3) - 6
       = 54 + 9a + 3b - 6
       = 48 + 9a + 3b
Since (x - 3) leaves remainder then
    48 + 9a +3b = 36
or 48 + 9( -10 - 2b/4) +3b = 36 [Using (i)]
or 48 -90 - 18b /4 + 3b =36
or 192 - 90 - 18b + 12b /4 = 36
or 102 - 6b = 36×4
or 102 - 6b =144
or -6b = 144 - 102
or -6b = 42
or b = -42/6
or b = -7

therefore a = -10 - 2b / 4
                  = -10 - 2(-7) / 4
                  = -10 + 14 /4
                  = 4/4
                  = 1
 
a = 1 and b = -7