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.



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