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If the G.C.D. of the polynomials x³−3x²+px+24 and x²−7x+q is (x−2), then the value of (p+q) is

A. 0
B. 20
C. -20
D. 40

1 Answer

3 votes

Final answer:

By applying the Remainder Theorem to the given polynomials with the factor (x-2), we find that p = -10 and q = 10. Adding these together, we get p + q = 0. Therefore, the correct option is A. 0.

Step-by-step explanation:

If the G.C.D. (Greatest Common Divisor) of the polynomials x³−3x²+px+24 and x²−7x+q is (x−2), and we want to find the value of (p+q), we must apply polynomial division or use the Remainder Theorem. Because (x−2) is a factor of both polynomials, when we substitute x = 2 into the polynomials, they should both equal zero. This means:

  • For the first polynomial: 2³−3(2)²+p(2)+24 = 0
  • For the second polynomial: 2²−7(2)+q = 0

Calculating these, we get:

  • 8∓12+2p+24 = 0 ⇒ 2p = −20 ⇒ p = −10
  • 4−14+q = 0 ⇒ q = 10

Add p and q together to find p + q = −10 + 10 = 0.

Therefore, the answer is A. 0.

User Raivo Kolde
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