How many real roots does f(x) = x^3+5x+1 have?

x^3 = odd polynomial = at least 1 real root
Can I follow Descartes' rule for this? No sign change, so there aren't any positive zeros.

when I sub in -x for x
(-x)^3+ 5(-x) + 1

= -x^3-5x+1

Then there is one sign change, so 1 negative root.

So my conclusion is that there is one real root. Is this right?

I agree. The Descartes rule says that there cannot be more than one real root in this case, and that it will be negative. There is another rule that says there must be at least one real root. Therefore the number of real roots is one.

Yes, you are correct. To determine the number of real roots for the polynomial f(x) = x^3+5x+1, you can use Descartes' rule of signs.

First, note that the exponents of the polynomial are all odd, which means that it is an odd polynomial and must have at least one real root.

To apply Descartes' rule, you need to examine the changes in sign of the coefficients within the polynomial.

For f(x) = x^3+5x+1, there is no sign change in the original polynomial (from x^3 to 5x), so there are no positive real roots.

Next, you can substitute -x for x in the polynomial, which gives you f(-x) = (-x)^3 + 5(-x) + 1 = -x^3 - 5x + 1. In this new polynomial, there is one sign change (from -x^3 to -5x), indicating that there is exactly one negative real root.

Therefore, your conclusion that f(x) = x^3+5x+1 has one real root is correct.