Question

The expression x3+px2+qx−3, where p and q are constants, has a factor of x−3 and leaves a remainder of 15 when divided by x+2.Find the values of p and of q.​

Answer 1

Answer:

-1

Step-by-step explanation:

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

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