Determine the real roots of the polynomial equation: (x^2+1)(x-4)=0

real roots? x=4.

imaginary roots? x=+-sqrt(-1)

ok thanks but how do you know if the roots are imaginary or not?

They are imaginary because a negative number cannot be square rooted.

ok thanks :D

To determine the real roots of the polynomial equation, we need to solve the equation (x^2+1)(x-4)=0.

First, we can recognize that this equation is obtained by multiplying two factors, (x^2+1) and (x-4), and setting the product equal to zero. According to the zero-product property, if a product of factors is equal to zero, then at least one of the factors must be equal to zero.

So, we set each factor equal to zero and solve for x:

(x^2+1) = 0
To solve this equation, we subtract 1 from both sides:
x^2 = -1

Since x^2 can't be negative for real numbers, there are no real solutions for this portion of the equation.

(x-4) = 0
To solve this equation, we add 4 to both sides:
x = 4

Therefore, the real root of the polynomial equation (x^2+1)(x-4)=0 is x = 4.