How do you use the rational root theorem to find the roots of #8x^3-3x^2+5x+15#?

2 Answers
Oct 20, 2015

Answer:

#(1,2,4,8)/(1,3,5,9,15,45)=1,1/3,1/5,1/9,1/15,1/45,2,2/3,2/5,2/9,2/15,2/45,4,4/3,4/5,4/9,4/15,4/45,8,8/3,8/5,8/9,8/15,8/45#

Explanation:

The general form for a polynomial is as follows:
#px^n+a_n-1+...a_1x+q#

The rational roots theorem states that, to find any potential zeroes of a given polynomial function, the formula is: #("factors of p")/("factors of q")#, where #p# and #q# are the first and last coefficients of the given function. In this case, the factors of #p# are 1,2,4,and 8, and the factors of q are 1,3,5,9,15,and 45.

Once you have all the potential zeroes of the function, you just have to test them using synthetic division.

Oct 20, 2015

Answer:

You can't.

You can determine that #8x^3-3x^2+5x+15=0# has one Real irrational root in #(-1, -3/4)# and no rational roots.

Explanation:

By the rational root theorem, any rational roots of #8x^3-3x^2+5x+15 = 0# must be expressible in lowest terms in the form #p/q# where #p, q in ZZ#, #q != 0#, #p# a divisor of the constant term #15# and #q# a divisor of the coefficient #8# of the leading term.

The prime factorisation of #8# is #2*2*2#.
The prime factorisation of #15# is #3*5#.

That means that the only possible rational roots are:

#+-1/8#, #+-1/4#, #+-3/8#, #+-1/2#, #+-5/8#, #+-3/4#, #+-1#, #+-5/4#, #+-3/2#, #+-15/8#, #+-5/2#, #+-3#, #+-15/4#, #+-5#, #+-15/2#, #+-15#

That's quite a lot of possibilities to try, so let's narrow it down.

Let #f(x) = 8x^3-3x^2+5x+15#

Then #f'(x) = 24x^2-6x+5#, which has a negative discriminant.

So we can deduce that #f(x)# is strictly monotonic increasing.

#f(-1) = -8-3-5+15 = -1#

#f(-3/4) = -8*3^3/4^3-3*3^2/4^2+5*3/4+15#

#= -27/8-27/16+15/4+15#

#= (-54-26+60+240)/16 = 220/16 = 55/4#

So #f(x) = 0# has one Real root in #(-1, -3/4)#, but it is not rational.