Solving a fourth degree equation Well, I want to find the roots of the equation:
$$x^4+6x^3-9x^2-162x-243=0\tag1$$
And Wolfram Alpha tells me that I can factor this and get:
$$(x^2-3x-9)(x^2+9x+27)=0\tag2$$
Then it will be easy to solve... But my question is, how can I go from $(1)$ to $(2)$?
 A: Let $(x^2+ax+b)(x^2+cx+d)=x^4+6x^3−9x^2−162x−243$.
Then $a+c=6, ac+b+d=-9, ad+bc=-162, bd=-243$.
You can try $a,b,c,d$ as integers, and since there are only finitely many possibilities, it is do-able.
Otherwise, by substituting $c=6-a$ and $d=-243/b$, you will get a quadratic equation of $b$ with coefficient being functions of $a$. This method is very messy bu do-able too.
In general, I don't think there is a clean way to do it. The formula is very messy too, so often when you encounter this as an exercise you can factor them in to easy ones (method 1: try integers!).
EDIT: Do take a look at the first comment below!
A: In case anyone ever looks at this question again...it is best to try things that might give a factorization first. What I notice is the powers of 3 in $x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \;$ This suggests taking $x=3t,$ then dividing through by the resulting constant factor $81,$ resulting in
$$ t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
The two things to try next are
$$ (t^2 + At + 1) (t^2 + B t - 3)  $$
and
$$ (t^2 + Ct - 1) (t^2 + D t + 3)  $$
The first one fails, you get $A = 8/5$ and $B = 2/5,$ the coefficient of $t^2$ is wrong.
The second one works, $C+D=2,$ then $3C-D= -6.$ Add these to get $4C = -4,$ then $C = -1,$ then $D = 3.$ The $t^2$ term also works out correctly. 
$$ (t^2 -t - 1) (t^2 + 3 t + 3) =   t^4 + 2 t^3 - t^2 - 6 t - 3 \; . \; $$
To get back to $x$ I suggest multiplying by $9 \cdot 9 = 81$ for
$$ (9t^2 -9t - 9) (9t^2 + 27 t + 27) =   81t^4 + 162 t^3 - 81 t^2 - 486 t - 243 \; . \; $$
Finally $t = x/3$ becomes
$$ (x^2 -3x - 9) (x^2 + 9 x + 27) =   x^4 + 6 x^3 - 9 x^2 - 162 x - 243 \; . \; $$
A: I would use a CASIO calculator:



*

*Type in the equation $X^4+6X^3-9X^2-162X-243$. To type "$X$", you must press ALPHA, then the button that has $X$ above it. In my calculator, $X$ is located above the close bracket ")", "SHIFT" is located on the top left, "ALPHA" is right next to "SHIFT".

*Press SHIFT, then CALC (SOLVE), then "$=$". When the first real result is out (for example, $X=-1.854101966$, press SHIFT and RCL (STO), then press $A$ to make a new variable $A=-1.854101966$.

*Type in the equation $\dfrac{X^4+6X^3-9X^2-162X-243}{X-A}$, then SHIFT, CALC (SOLVE) in that order then press "$=$" two times. The calculator will automatically calculate another real root, this cannot be $A$ because that violates the defined condition. Press SHIFT, RCL (STO), $B$ in that order to get the second variable, in this case $B=4.854101966$. I have tried this and it takes quite a while, about over a minute to get the result.

*Now use the calculator again to calculate the sum and product of two variables:
${\begin{cases}S=A+B=3\\P=AB=-9\end{cases}}$
Using the Vieta's formula, we know that $A$ and $B$ are two real roots of the equation $x^2-3x-9=0$, so when we write the first equation as a product of two quadratic equations, one of them should be $x^2-3x-9$. 
$x^2+9x+27$ does not have real roots however, so the calculator will not give the roots of that equation. Instead, using a method to divide $x^4+6x^3-9x^2-162x-243$ by $x^2-3x-9$ to get $x^2+9x+27$.
A: A straightforward method that reduces the problem to a standard Rational Roots Theorem problem, is given here. We start with reducing the polynomial modulo the polynomial $x^2 - p x - q$, this yields:
$$(p^3+6 p^2+2 p q-9 p+6 q-162) x + p^2 q+6 p q+q^2-9 q-243 \tag{1}$$
Then we need to find integers $p$ and $q$ such that this expression is identical zero. Putting $x = -\frac{q}{p}$ and solving for $p$ yields:
$$p = -\frac{6 q\left(q-27\right)}{q^2+243}\tag{2}$$
This then sets one linear combination of the two coefficents in (1) zero. We need to set another one to zero. We can take the constant term and use use (2) s to eliminate $p$. This yields:
$$q^6-9 q^5-729 q^4+13122 q^3+177147 q^2-531441 q-14348907 = 0\tag{3}$$
The problem has now become a straightforward application of the Rational Roots Theorem. The constant term is $-14348907 = -3^{15}$ So, we know that $q$ has to be a power of $3$ up to a sign. If we substitute $q = 2 + t$, then we get a polynomial equation in $t$ which has a constant term equal to the r.h.s. of (3) for $q = 2$, which is $-14610113$ this has prime factors of $7$, $29$ and $71971$. Adding 2 to these possible solutions yields the corresponding value for $q$, but this has to be a power of $3$ up to a sign. This means that $q=9$ and $q = -27$ are the only possible solutions.
Inserting $q = 9$ in (2) gives $p = 3$, while inserting $q = -27$ in (2) gives $p = -9$. The quadratic factors $x^2 - p x - q$ are thus $x^2 -3 x - 9$ and $x^2 + 9 x + 27$
A: To solve $x^4+6x^3−9x^2−162^x−243=0$,
first try and factor, if possible: $(x^2−3x−9)(x^2+9x+27)=0$
So since this factors into to quadratics, we can use quadratic formula and some other simplification methods such as completing the square.
The roots of $x^2-3x-9$ are the same as the roots of $x^2-x-1$, multiplied by three since the absolute value of coefficients strictly increases by a power of $3$. This all amounts to $(3/2)(1±√5)$ being the roots of $x^2-3x-9$, and as for $x^2+9x+27$, one notices the increasing powers of $9$, and therefore a root of $x^2+9x+27$ is equivalent to a root of $x^2+x+1$, multiplied by nine. This amounts to $(9/2)(-1±i√3)$ being the roots of $x^2+9x+27$. 
The complete list of solutions to 
$x^4+6x^3−9x^2−162^x−243=0$
is
$(9/2)(-1+i√3)$
$(9/2)(-1-i√3)$
$(3/2)(1+√5)$
$(3/2)(1-√5)$
