# 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 = • 7

if we have to find the value of x and each term of the equation is in multiplication, then equate each term to the given value on the equal to sign.

x+2=2600          3x+4=2600          3x+7=2600          x+3=2600

x=1300               x= 2596/3            x=2593/3              x=2597

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