list all the possible rational roots of 6x4-2x3+5x2+x-10=0
always try x = 1 first
6-2+5+1-10 = 0 sure enough
so x = +1 is a root
and
(x-1) is a factor
so divide the mess by (x-1)
get
6 x^3 + 4 x^2 + 4 x +10
I can only find one real root of that, at around x = -1.23
so I suppose you could divide
6 x^3 + 4 x^2 + 4 x +10
by
(x+1.23)
and get a quadratic
that quadratic will not have any real roots, but will have two complex roots, complex conjugates.
To find the possible rational roots of a polynomial equation, we can use the Rational Root Theorem. According to this theorem, any rational root of a polynomial equation can be expressed as a fraction in the form of p/q, where p is a factor of the constant term (the last term in the equation) and q is a factor of the leading coefficient (the coefficient of the highest power term).
In your equation, 6x^4 - 2x^3 + 5x^2 + x - 10 = 0, the constant term is -10, and the leading coefficient is 6.
First, let's find the factors of the constant term (-10):
-10 can be factored as 1 * -10, -1 * 10, 2 * -5, or -2 * 5.
Next, let's find the factors of the leading coefficient (6):
6 can be factored as 1 * 6, -1 * -6, 2 * 3, or -2 * -3.
Now, we can form all possible fractions using the factors we obtained:
p/q = ±1, ±2, ±5, ±10 divided by ±1, ±2, ±3, ±6.
Combining all the factors, we have the following possible rational roots:
±1, ±2, ±5, ±10, ±1/2, ±5/2, ±1/3, ±5/3, ±1/6, and ±5/6.
So, the possible rational roots of the equation 6x^4 - 2x^3 + 5x^2 + x - 10 = 0 are ±1, ±2, ±5, ±10, ±1/2, ±5/2, ±1/3, ±5/3, ±1/6, and ±5/6.