using rational root theorem to list possible rational roots for x³ + x² - 3x - 6

Question

Using the rational root theorem to list the possible rational roots for the polynomial equation

Answer 1

I agree with Kirill's determination of the answer. A quick check of easy possibilities (for example, +/- 1 or 2) confirms that x = 2 is a root. Given this information, you can say that (x - 2) is a factor of the cubic equation and factor out that root. This is accomplished by:

 

(x-2)(ax²+bx+c) = x³ + x² -3x - 6

 

In order to solve for the 3 coefficients (a,b,c), you can use several methods. Synthetic division is a good method to back into the remaining quadratic equation. Now you have:

 

(x-2)(x² + 3x + 3) = 0

 

The only step left is to confirm whether or not the roots of the quadratic portion are real or complex. This step is accomplished by utilization of the quadratic formula by checking the sign of the determinant. The determinant term is:

 

b² - 4ac = 3² - 4(1)(3) = 9 - 12 = -3 (which is less than zero, an indication of a complex root and not a real root).

 

As such, the only rational root for the original cubic equation is x = 2. The other roots are complex and are -3(1+i)/2 and -3(1-i)/2. 

Answer 2

It's easier to understand and remember how to use the rational root theorem if you see where it comes from by considering the roots of a polynomial after it's been fully factored. If you fully factor a polynomial then you get a bunch of linear terms multiplied by each other. Just like a quadratic can be factored into two binomials multiplied by each other, a cubic can be factored into three binomials:

 

(Ax+B)(Cx+D)(Ex+F)=0.

 

Because of the zero multiplication property, this factored cubic has the three roots x=-B/A, -D/C, and -F/E. For roots where both capital letters are positive or negative integers, the roots are rational numbers. But those capital letters can also be irrationals or they can even have imaginary parts. 

 

You can multiply everything out in the equation above, first by using FOIL on the first two terms to get a trinomial and then by multiplying that trinomial by (Ex+F) to get the fully expanded cubic:

 

ACEx^3 +(complicated expression #1)(x^2) + (complicated expression #2)(x) + BDF = 0

 

The rational root theorem comes from the relationship between the roots (-B/A, -D/C, and -F/E)  and the terms in the cubic equation. Find all positive and negative factors of the leading coefficient. Those are the possible denominators of rational roots. Do the same thing with the units term. Those are the possible numerators of rational roots.

 

For the cubic you gave, the leading coefficient is 1. In the general equation above, ACE =1, so the possible denominators of rational roots must be ±1. The units coefficient is –6, so in the general equation above, BDF=-6, so the possible numerators of rational roots must be ±1, ±2, ±3, or ±6.

 

Consider all possible combinations of numerators and denominators, and you have:

±1/1, ±2/1, ±3/1, or ±6/1

or

±1, ±2, ±3, or ±6.

Answer 3

First, it is immediately obvious that x=2 is the root. So x³+x²-3x-6 can be written as follows:

 

x³+x²-3x-6=(x-2)(ax²+bx+c)

 

Coefficients a, b, and c can be determined by doing synthetic division of original polynomial by (x-2).

 

(x³+x²-3x-6)/(x-2)=(x²+3x+3)

 

The determinant of quadratic trinomial x²+3x+3 is D=9-3*4=-3<0, therefore equation x²+3x+3=0 does not have roots in the field of real numbers. Any polynomial equation with rational coefficients must have even number of complex roots, therefore original polynomial has only one rational root, x=2.