Question

(x+1)(2x+3)(2x+5)(x+3)=945

Answer 1

factors of 945 are 3 , 3 , 3 , 5 , 7, 1

Here x = 2
x + 1 = 2 + 1 = 3
2x + 3 = 2×2 + 3 = 7
2x + 5 = 2×2 + 5 = 9
x + 3 = 2 + 3 = 5
Clearly 3 , 7 , 9 and 5 are the factors of 945...if you multiply 3 , 7 , 9 , 5 together you get 945..

( x + 1 ) ( 2x + 1 ) ( 2x + 5 ) ( x + 3 ) = 945
=>( 2 + 1 ) (2×2+1) (2×2+5) (2 + 3) = 945
Thus x = 2

Answer 2

(x+1)(2x+3)(2x+5)(x+3)=945

(2x^2 +5x+ 3)(2x^2+11x+15) =945

4x^4+32x^3+ 91x^2 + 108x +45=945

4x^2 + 32x^3+ 91x^2 +108x-900=0

4x^4 - 8x^3 +40x^3 -80x^2 +171x^2 -342x+450x- 900=0

4x^3( x-2) + 40x^2(x-2) +171x(x-2) +450(x-2)=0

4x^3 +40x^2 + 171x +450)(x-2) =0

4x^3+ 40 x^2 +171 x + 450)(x-2) =0

4x^3 +24x^2 +16x^2 +96x+75 x +450) (x-2) =0

[4x^2(x+6) +16x( x+6)+ 75(x+6)] (x-2)=0

(4x^2+16x +75)(x+6)( x-2)=0

FROM THIS WE CAN GET TWO SOLUTIONS FOR X= -6, 2....

LET US SOLVE 4x^2+ 16x+75=0 USING QUADRATIC FORMULA....

X= -16±√256-1200÷8 = -16±√-944÷8

x=-4±√59i ÷2

THUS, X= -6, 2, -4±√59i÷2

Hope it is helpful......