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)
list all the possible rational roots of 6x4-2x3+5x2+x-10=0
2 answers
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.
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.
1 answer