Foolproof method for simplifying polynomials with four terms?

Question

When simplifying quadratic equations you have two options:

• factoring (which may or may not work)
• or the quadratic formula (which will always find the answer)

For quadrinomials what is the go to method?
For example if you try to factor this polynomial

$x^3 + 2x^2 + x - 4$

You would end up with:

$x^2(x + 2) + 1(x - 4)$

There must be a different method to solve this equation. Is there a quadrinomial equivalent to the quadratic formula?

Answer 1

A previous answer correctly showed that $\quad x^3 + 2x^2 + x - 4=(x - 1) (x^2 + 3 x + 4). \quad $ and this is apparent immediately if you simply substitute $x=1$ into the polynomial.

Another method is using the cubic formula

\begin{equation} n=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a} \end{equation} where $\qquad a=1\qquad b=2\qquad c=1\qquad d=-4$

In some situations, this leads to transient complex values but robust software like WolframAlpha can handle it is you need a particular solution.

Answer 2

We put $f(x) = x^3+2x^2 +x-4$

we can see$: f(1)=0$ $ \Rightarrow$ $ f(x) $ is divisible by$ (x-1)$

After division :

$f(x) = (x-1)(x^2 +3x+4 )$

Answer 3

There is a linear substitution that turns

$$ax^3+bx^2+cx+d=0$$ to one of $$4t^3\pm3t=y$$

(set $x=pt+q$ and identify, to a constant coefficient; tedious but doable).

Then use one of

$$t=\cos u, t=\cosh u, t=\sin h$$

to obtain one of

$$\cos 3u=y,\cosh3u=y,\sinh3u=y.$$

Solve for $3u$, then $u$, then $t$, then $x$.