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.
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It doesn't factor over the rationals. You'd need the cubic equation. [Note are you sure the coefficients are exactly what you wrote? A slight change may make it factorable.] – coffeemath Jul 10 '22 at 14:54
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At all integer values of x, it's even. – Roddy MacPhee Jul 10 '22 at 14:55
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No error in the coefficients – pravakar wang Jul 10 '22 at 15:00
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1Perhaps $x^3+6x^2+13x+8$? This has a rational root. – Dietrich Burde Jul 10 '22 at 15:20
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@DietrichBurde could you please check the answer submitted by me. – Janaka Rodrigo Jul 10 '22 at 17:57
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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:
Dietrich Burde
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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
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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.
Janaka Rodrigo
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I believe the method used here is correct and simplifications too. Since the values you get for a³ are in surd form (irrationals) and you need to get cube root of them, numerical form is much complicated , therefore I have not provided the final values. I hope given guidance may be enough to solve the problem. Your suggestions are greatly appreciated. – Janaka Rodrigo Jul 11 '22 at 02:09
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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).
Oscar Lanzi
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