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in Complex number and Quadratic equations by (46.6k points)

Let p(x) be a quadratic polynomial such that p(0) = (1). If p(x) leaves remainder 4 when divided by x - 1 and it leaves remainder 6 when divided by x + 1, then:

(A)  p(-2) = 11

(B)  p(2) = 11

(C)  p(2) = 19

(D)  p(-2) = 19

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Best answer

Correct option  (D) p(-2) = 19

 Let us consider 

p(x) = Ax2 +  Bx + C

Substituting x = 0 in this equation, we get

p(0) = 0 + 0 + C

1 = C     ............(1)

Also, when p(x) leaves remainder 4 when divided by x - 1, we get

x - 1 = 0 ⇒ x = 1

Therefore,

p(1) = A + B + C ⇒ 4 = A + B + C  .....(2)

Substituting C = 1 in Eq. (2), we get

4 = A + B + 1 ⇒ A + B = 3  .....(3)

Also, when p(x) leaves remainder 6 when divided by x + 1, we get

x + 1 = 0 ⇒ x = -1

⇒ p(-1) = A - B + C

⇒ 6 = A - B + C  [C = 1 from Eq. (1)]

6 = A - B + 1 ⇒ A - B = 5     .....(4)

Solving Eqs. (3) and (4), we get

A + B = 3

A - B = 5

Eq. (3) + Eq. (4): 2A = 8 ⇒ A = 4  ....... (5)

Eq. (3) - Eq. (4): 2B = -2 ⇒ B = 1  .........(6)

Substituting C = 1, A = 4, B = -1 from equation (1), (5) and (6) in Eq. (I), we get 

p(x) = 4x2 - x + 1

p(-2) = 4(-2)2 - (-2) + 1

= 4 × 4 + 2 + 1 = 16 + 2 + 1 ⇒ p(-2) = 19

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