How do you use the rational root theorem to find the roots of #5x^4 + 9x^3 + 5x^2 + 2x + 4 = 0#?

2 Answers
Jul 4, 2015

Answer:

List the possible rational roots and then test them. None of them are roots.

Explanation:

For #5x^4+9x^3+5x^2+2x+4 = 0#, we have leading coefficient #5# and constant #4#.
The divisors of #5# are #1# and #5#.
The divisors of #4# are #1#, #2#, and #4#.

So,the possible rational roots are:

#1, 2, 4, 1/5, 2/5, 4/5, -1, -2,-4, -1/5, -2/5, -4/5#

Since all of the coefficients are positive, there can be no positive roots.

Testing by synthetic division we can find that #-1# is not a root and for #-2#, it is not a root and we get alternating signs, so #-2# is a lower bound on the roots,

So we need to test #-1/5, -2/5, -4/5#, none of which are roots.

The equation has no rational roots.

Jul 4, 2015

Answer:

Any rational roots must be #+-4,+-2,+-1,+-4/5,+-2/5# or #+-1/5#

None of these are roots, so it has no rational roots.

Explanation:

Let #f(x) = 5x^4+9x^3+5x^2+2x+4#

By the rational root theorem, the rational roots of #f(x) = 0# must all be of the form #p/q# in lowest terms with #p# a divisor of #4# and #q# a divisor of #5#.

So the only possible rational roots are #+-4,+-2,+-1,+-4/5,+-2/5# or #+-1/5#.

None of these are roots, so there are no rational roots.

In fact #f(x) = 0# has no real roots.

graph{5x^4+9x^3+5x^2+2x+4 [-11.99, 11.42, -1.56, 10.15]}