If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder (-3/4) -Maths 9th

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Given, (9x2 + 3px + 6q), when divided by (3x + 1) leaves a remainder −34−34 ∴ f(x) = 9x2 + 3px + 6q – (−34)(−34) = (9x2+3px+6q+34)(9x2+3px+6q+34) is exactly divisible by (3x + 1)  ∴ f(−34)(−34) = 0 ⇒ 9(−34)(−34)2 + 3 p . (−34)(−34) + 6q + 3434 = 0 ⇒ 6q - p + 7474 0  ⇒ 24q - 4p + 7 = 0                      ......(i) Now, the expression g(x) = qx2 + 4px + 7 is exactly divisible by x + 1  ⇒ g(–1) = 0 ⇒ q – 4p + 7 = 0                     ...(ii)  Solving equations (i) and (ii), we get q = 0, p = 7474.

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