To Factorise $x^3 +6x^2+13x +4$

Question

I have tried factorisation by trial using factor theorem, to no avail. The expression is neither cyclic nor reciprocal. I have also tried arrangement and grouping of terms. I have run out of ideas to factorise this expression. Please help me out. Thanks.

Answer 1

It is easy to see that the polynomial has no root over the finite field $\Bbb F_5$. So it is irreducible in $\Bbb F_5[x]$ and hence also in $\Bbb Z[x]$ and $\Bbb Q[x]$. So there is no way to factor it.

Reference:

On the irreducibility of a polynomial and Gauss lemma

Answer 2

Since $\pm 1,\pm 2,\pm 4$ aren't roots of the equation, the polinomial is irreducible over rationals. You need the cubic root formula if you want to compute the roots

Answer 3

Let's consider the cubic equation,
x³+ 6x² + 13x + 4 = 0 terms can be rearranged to get,
(x+2)³ + (x+2) = 6 and compared with the identity , (a + b )³ -3ab(a+b)= a³ + b³,
Now you get x+2 = a+b, -3ab = 1 and a³ + b³ = 6,
Last two equations you can solve to find a and b , and finally by the first equation you can find x .
Now you can obtain factors of the given cubic polynomial.
Added this for your more convenient ,
When you solve two equations of a, b by eliminating b , you get quadratic equation of a³ and you can use quadratic formula to find two roots of a³ , now by the symmetry of equations of a,b you can say one root is for a³ and the other root is for b³. Then by getting the cube root of those values you can find a,b and you have only one real root a+b and a linear factor is given by x +2 -(a+b). By equating the coefficients you can obtain the other factor that is a quadratic .
To verify you can differentiate the polynomial function with respect to x and the derivative can not be zero because the discriminant of the quadratic equation is negative. Therefore by the shape of the graph there should be only one linear factor.

Answer 4

We can disprove all rational roots with minimum trials by girst noting that the cubic and quadratic terms match

$(x+2)^3=x^3+6x^2+12x+8.$

So with $y=x+2$ we have

$x^3+6x^2+13x+4=y^3+x-4=y^3+y-6.$

Then any real root for $x^3+6x^2+13x+4=0$ must satisfy $x<0$ and $y=x+2>0$, and the only candidate allowed by the Rational Root Theorem which satisfies this inequality is $-1$ (which is easily seen to fail).