Respuesta :
Hello,
(x²-1)(x²+2)(x+3)(x-4)(x+1)=0
==>(x+1)(x-1)(x-i√2)(x+i√2)(x+3)(x-4)(x+1)=0
Roots are -1 double, 1, i√2,-i√2,-3 and 4.
(x²-1)(x²+2)(x+3)(x-4)(x+1)=0
==>(x+1)(x-1)(x-i√2)(x+i√2)(x+3)(x-4)(x+1)=0
Roots are -1 double, 1, i√2,-i√2,-3 and 4.
Answer: The roots of the given equation are [tex]1,~-1,~\sqrt2i,~-\sqrt2i,~-3,~4,~-1.[/tex]
Step-by-step explanation: We are given to solve the following equation by identifying all of its roots including any imaginary numbers and multiple roots :
[tex](x^2-1)(x^2+2)(x+3)(x-4)(x+1)=0~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Since the given equation is in factored form, so the roots of the equation will be found by equating each factor to zero.
Therefore, the roots of the given equation (i) are
[tex]x^2-1=0\\\\\Rightarrow x^2=1\\\\\Rightarrow x=\pm\sqrt1~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=\pm 1,\\\\\\x^2+2=0\\\\\Rightarrow x^2=-2\\\\\Rightarrow x=\pm\sqrt{-2}~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=\pm\sqrt2i,\\\\\\x+3=0\\\\\Rightarrow x=-3,\\\\\\x-4=0\\\\\Rightarrow x=4\\\\\\\textup{and}\\\\\\x+1=0\\\\\Rightarrow x=-1.[/tex]
Thus, the roots of the given equation are [tex]1,~-1,~\sqrt2i,~-\sqrt2i,~-3,~4,~-1.[/tex]