InterviewSolution
Saved Bookmarks
| 1. |
If ax3 + bx2 + x – 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x – 2), find the values of a and b. |
|
Answer» Let p(x) = ax3 + bx2 + x – 6 (x + 2) is a factor of p(x) ⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2] ⇒ a(-2)3 + b(-2)2 + (-2) – 6 = 0 ⇒ -8a + 4b – 2 – 6 = 0 ⇒ -8a + 4b = 8 ⇒ -2a + b = 2 …(i) It is given that p(x) leaves the remainder 4 when it divided by (x – 2) i.e. p(2) = 4. ⇒ a(2)3 + b(2)2 + (2) – 6 = 4 ⇒ 8a + 4b – 4 = 4 ⇒ 8a + 46 = 8 ⇒ 2a + b = 2 …(ii) Solving (i) and (ii), we get a = 0 and b = 2. |
|