Question

Find all real and imaginary solutions to the equation.
x^3-2x^2+16x-32=0

Answer 1

The solution of
`x^3-2x^2+16x-32=0` is
`\mathbf{x=2,x=4i,x=-4i}`

So, the real solutions is: x = 2

Imaginary solutions are : x = 4i, x = -4i

Explanation:

We need to find real and imaginary solutions to the equations
`x^3-2x^2+16x-32=0`

First of all we will make groups

`(x^3-2x^2)(+16x-32)=0`

Finding common terms from the groups

`x^2(x-2)+16(x-2)=0\\(x-2)(x^2+16)=0`

Now we know that if ab=0 then a=0 , b=0

`x-2=0, x^2+16=0\\Simplifying:\\x=2, x^2=-16\\Taking\: square\: root\\x=2,√(x^2)=√(-16)\\x=2, x=\pm√(-16)\\x=2,x=\m√(-1)√(16) \\We\:know\:√(-1)=i \:and √(x) \:√(16)=4 \\x=2,x=\pm4i\\x=2,x=4i,x=-4i`

The solution of
`x^3-2x^2+16x-32=0` is
`\mathbf{x=2,x=4i,x=-4i}`

So, the real solutions is: x = 2

Imaginary solutions are : x = 4i, x = -4i