Question

(x+1)(x+3)(x-5)(x-7)=192

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

(x+1)(x+3)(x-5)(x-7)=192

$x(x + 3) + 1(x + 3)[(x - 5)(x - 7)]= 192$

${x}^{2} + 3 x + x + 3[(x - 5)(x - 7)] =192$

${x}^{2} + 4x + 3 + x(x - 7) - 5(x - 7)=192$

${x}^{2} + 4x + 3 + {x}^{2} - 7x - 5x + 35 = 192$

$= > 2 {x}^{2} + 4x - 7x - 5x + 3 + 35 = 192$

$= > 2 {x}^{2} - 8x + 38 - 192 = 0$

$= > 2 {x}^{2} - 8x + 154 = 0$

Taking 2 common from equation :-

$= > {x}^{2} - 4x + 77 = 0$

$= > {x}^{2} - 11x + 7x + 77 = 0$

$= > x(x - 11) - 7(x - 11) = 0$

$= > (x - 7)(x - 11) = 0$

=>x-7=0

${\boxed{x = 7}}$

=>x-11=0

${\boxed{x = 11}}$

$\therefore{\bold{Roots \:are\: x=7\& \:x=11}}$