Question

If ax^3+bx^2+x-6 has (x+2) as
a factor and leaves a remainder 4 when divided by (x-2),find the value of 'a' and 'b'.

ANSWER
a=0
b=2

Answer 1

Given f(x) = ax^3 + bx^2 + x -6.

Given that (x + 2) is a factor of f(x).

= > x + 2 = 0

x = -2.

Plug x = -2 in f(x), we get

= > f(-2) = a(-2)^3 + b(-2)^2 + (-2) - 6 = 0

              = -8a + 4b - 2 - 6 = 0

              = -8a + 4b - 8 = 0

              = 2a - b = -2 ----- (1)



Given that x - 2 leaves a remainder 4.

= > x - 2 = 0

= > x = 2

plug x = 2 in f(x), we get

= > 4 = a(2)^3 + b(2)^2 + (2) - 6

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

= > 4 = 8a + 4b - 4

= > 8a + 4b = 8

= > 2a + b = 2  ----- (2)



On solving (1) & (2), we get

2a - b = -2

2a + b = 2

-----------------

  4a = 0

     a = 0


Substitute a = 0 in (1), we get

= > 2a - b = -2

= > 2(0) - b = -2

= > -b = -2

= > b = 2.



Therefore a = 0 and b = 2.



Hope this helps!