Question

(×-5)(×-3)(×₊1)(×₊3)-(ײ-2×-9)²₊36

Answer 1

$\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(x+3\right)-\left(x^2-2x-9\right)^2+36=0$

$\left(x^2-2x-9\right)^2=\left(x^2-2x-9\right)\left(x^2-2x-9\right)$

$=\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(x+3\right)-\left(x^2-2x-9\right)\left(x^2-2x-9\right)+36$

$\left(x^3-7x^2+7x+15\right)\left(x+3\right):\quad x^4-4x^3-14x^2+36x+45$

$-\left(x^2-2x-9\right)\left(x^2-2x-9\right):\quad -x^4+4x^3+14x^2-36x-81$

$x^4-4x^3-14x^2+36x+45-x^4+4x^3+14x^2-36x-81+36:\quad$

x=0