How do you use the rational root theorem to find the roots of #3x^4 + 5x^3 + 3x^2 - 7x + 9 = 0#?

2 Answers
Jul 22, 2015

Answer:

There are no rational roots.

Explanation:

Rational roots would be the factors of #9/3# that is 3. These can be 1,-1,3,-3. None of these values satisfy the given equation, hence it is concluded that there are no rational roots.

Jul 22, 2015

Answer:

From the rational root theorem we find that the only possible rational roots are #+-1/3#, #+-1#, #+-3# and #+-9#.

None of these is a root, so the equation has no rational roots.

Explanation:

By the rational root theorem, any rational root of #3x^4+5x^3+3x^2-7x+9=0# must be of the form #p/q# where #p# and #q# are integers, #q != 0#, #p# a divisor of the constant term #9# and #q# a divisor of the coefficient #3# of the term #3x^4# of highest degree.

That means that the only possible rational roots are:

#+-1/3#, #+-1#, #+-3# and #+-9#

Let #f(x) = 3x^4+5x^3+3x^2-7x+9#

Then:

#f(1/3) = 65/9#

#f(-1/3) = 311/27#

#f(1) = 13#

#f(-1) = 17#

#f(3) = 393#

#f(-3) = 165#

#f(9) = 23517#

#f(-9) = 16353#

So none of the possible rational roots is a root of #f(x) = 0#

So #f(x) = 0# has no rational roots.

In fact it has no real roots at all:

graph{3x^4+5x^3+3x^2-7x+9 [-20, 20, -4, 16]}