If (x3 + ax2 + bx + 6) has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.
Given, (x3 + ax2 + bx + 6) exactly divisible by (x – 2)
⸫ x = 2 is a root of the above equation.
⇒ 23 + a (2)2 + b (2) + 6 = 0
⇒ 8 + 4a + 2b + 6 = 0
⸫ 4a + 2b = -14 …. (i)
Given, (x3 + ax2 + bx + 6) divided by (x – 3) leaves a remainder 3
⸫ 33 + a (3)2 + b (3) + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⸫ 9a + 3b = -30 …. (ii)
Solving simultaneously eq (i) and eq (ii), we get
a = -3, b = -1