Answer:
The solution of [tex]x^3-2x^2+16x-32=0[/tex] is [tex]\mathbf{x=2,x=4i,x=-4i}[/tex]
So, the real solutions is: x = 2
Imaginary solutions are : x = 4i, x = -4i
Step-by-step explanation:
We need to find real and imaginary solutions to the equations [tex]x^3-2x^2+16x-32=0[/tex]
First of all we will make groups
[tex](x^3-2x^2)(+16x-32)=0[/tex]
Finding common terms from the groups
[tex]x^2(x-2)+16(x-2)=0\\(x-2)(x^2+16)=0[/tex]
Now we know that if ab=0 then a=0 , b=0
[tex]x-2=0, x^2+16=0\\Simplifying:\\x=2, x^2=-16\\Taking\: square\: root\\x=2,\sqrt{x^2}=\sqrt{-16}\\x=2, x=\pm\sqrt{-16}\\x=2,x=\m\sqrt{-1}\sqrt{16} \\We\:know\:\sqrt{-1}=i \:and \sqrt{x} \:\sqrt{16}=4 \\x=2,x=\pm4i\\x=2,x=4i,x=-4i[/tex]
The solution of [tex]x^3-2x^2+16x-32=0[/tex] is [tex]\mathbf{x=2,x=4i,x=-4i}[/tex]
So, the real solutions is: x = 2
Imaginary solutions are : x = 4i, x = -4i
