solve the equation (x+2)(3x+4)(3x+7)(x+3) =2600

Given, (x+2)(3x+4)(3x+7)(x+3) = 2600

⇒ (x+2)(x+3)(3x+4)(3x+7) = 2600

⇒ (x² + 5x + 6)(9x² + 33x + 28) = 2600

Again multiplying the terms in the brackets on LHS, we get

9x⁴ + 78x³ + 247x² + 338x + 168 = 2600

⇒ 9x⁴ + 78x³ + 247x² + 338x - 2432 = 0 ... (1)

Now, on putting x = 2, equation (1) is equal to zero.

So, x-2 is a zero of (1).

∴  9x⁴ + 78x³ + 247x² + 338x - 2432 = (x - 2)(9x³ + 96x² + 439x + 1216)

Again factorizing the term (9x³ + 96x² + 439x + 1216) we get,

 (9x³ + 96x² + 439x + 1216) = ( 3x + 19 )( 3x² + 13x + 64)

So, 9x⁴ + 78x³ + 247x² + 338x - 2432 = (x - 2)( 3x + 19 )( 3x² + 13x + 64) = 0  [from (1)]

The expression ( 3x² + 13x + 64) can't be factorized further as it has no rational zeros.

So, Either (x-2) = 0 or (3x + 19) = 0

⇒ x = 2 or  x =