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

Question

If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder \(\big(-\frac{3}{4}\big)\) and qx² + 4px + 7 is exactly divisible by (x + 1), then the values of p and q respectively will be : 

(a) 0, \(\frac{7}{4}\)

(b) \(-\frac{7}{4}, 0\)

(c) Same 

(d) \(\frac{7}{4}, 0\)

Answer 1

 (d) \(\frac{7}{4}, 0\)

Given, (9x² + 3px + 6q), when divided by (3x + 1) leaves a remainder \(-\frac{3}{4}\)

∴ f(x) = 9x² + 3px + 6q – \(\big(-\frac{3}{4}\big)\) = \(\big(9x^2+3px+6q+\frac{3}{4}\big)\)

is exactly divisible by (3x + 1) 

∴ f\(\big(-\frac{3}{4}\big)\) = 0 ⇒ 9\(\big(-\frac{3}{4}\big)\)² + 3 p . \(\big(-\frac{3}{4}\big)\) + 6q + \(\frac{3}{4}\) = 0

⇒ 6q - p + \(\frac{7}{4}\) 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 = \(\frac{7}{4}\).