if x3+px2+qx+6 leaves remainder 3 when divided by x-3 and leaves remainder 0 when divided by x-2. find the values of p and q.
If the given polynomial is f(x) = x3 + px2 + qx + 6 then solution will follows as:
When f(x) is divided by x - 3 and x - 2, the remainders are 3 and 0 respectively.
∴ f(3) = 3 and f(2) = 0
⇒ (3)3 + p(3)2 + q(3) + 6 = 3 and (2)3 + p(2)2 + q(2) + 6 = 0
⇒ 27+ 9p + 3q + 6 = 3 and 8 + 4p + 2q + 6 = 0
⇒ 9p + 3q + 33 = 3 and 4p + 2q + 14 = 0
⇒ 9p + 3q = -30 and 4p + 2q = -14
⇒ 3p + q = -10 ... (1) and 2p + q = -7 .... (2)
On subtracting (1) and (2), we get
3p + q - (2p + q) = -10 -(-7)
⇒ p = -10 + 7 = -3
On putting p in (2), we get
2(-3) + q = -7
⇒ q = -7 + 6 = -1