Using the rational root theorem, find the solutions of the equation x^3 - 6x - 4 = 0.

Question

I got as far as (x+2)(x^2-2x-2). I see that the answer in the back of my book is -2, 1 +/- sq(3). (the sq(3) meaning the square root of 3). I get how I get the -2, but how do we get the answer of 1 +/- sq(3) from (x^2-2x-2)?

Answer 1

The rational root theorem gives all the possible rational roots (integer divided by an integer).  The possible rational roots are all factors of p (-4, the constant) divided by all factors of q (1, the coef on the high order term) both positive and negative.  Further, Descartes' rule of signs says there are either 0 or 2 positive rational roots.   So, the possible positive rational roots are 1, 2, and 4.  None of those work, so there are no positive RATIONAL roots.  If we look at -1, -2, and -4; only -2 works, so it is the only rational root.  If you divide the equation by (x+2, which is x minus the root), that leaves you with a quadratic equation, which can be solved by the quadratic formula.  This results in the other 2 non-rational roots (1+sqrt(3), 1-sqrt(3)).

Ken L.

Answer 2

Hello Alyianna.

Just by trials you can see that -2 is a solution of equation you were assigned. In fact,

(-2)³-6*(-2) - 4 = 0.

Then, you can divide the polynomials you were assigned by (x-(-2) = x+2, using the appropriate rule to divide a polynomial by a binomial:

x³ +0x²-6x-4: (x+2) = x²-2x-2 

which is where you got to.  So, now you need to solve  x^₂-2x-2 = 0 which is a quadratic equation. I assume you are familiar with the formula that uses the discriminant. In that case, the solutions of this equation are given by:

x = (+2 ±√(4-4(1)(-2))/(2*(1)) = (+2 ± 2√3)/2 = 1± √3. 

as you were asked to prove. Another possibility is to use the complete the square technique, but in general the discriminant formula is pretty straightforward to apply.

The rational root theorem only provides you with the rational roots of this equation if there are any. If you use the rational root theorem, remember that the solutions have the form p/q where

- q divides the coefficient of x³, 1,  and p divides the constant, 4. 

so, q can be only ±1 and p can be ±1, ±2, ±4,  and therefore p/q can take any of the values 1,-1, 2,-2, 4, -4 which you need to check one by one and you find that only -2 is actually a solution.  Then you need to use another method as the remaining two solutions are not rational numbers.

 

Hope this helps.

Cheers,

Maurizio