Factoring $a^4+2a^3 b-3a^2 b^2-4a b^3-b^4$

Question

I have tried rearranging and grouping, considering the expression $f(a)$ and finding a root in terms of $b$ (which so far have worked for this sort of problems), but none was fruitful. In fact, this will factor into two trinomials, but I am absolutely clueless so as to how to go about factoring it intuitively (without knowing in the first place that the factors are trinomials).

Answer 1

We know $ x^4 + 2 x^3 - 3 x^2 - 4 x - 1$ has no rational root, so possible factoring over the rationals is $$ (x^2 + A x - 1) (x^2 + B x + 1) $$ to get the $-1$ product as the constant term. Then cubic $A+B = 2,$ linear term $A - B = -4,$ so $2A = -2$ and $A = -1,$ then $B = 3.$ Then we check at the quadratic term, $AB + 1 - 1 = AB = -3,$ so it works.

$$ x^4 + 2 x^3 - 3 x^2 - 4 x - 1 = (x^2 - x - 1)(x^2 + 3x + 1) $$ Note how both quadratics have discriminant $5$

Another method that seems possible is to write it as a sum or difference of squares, perhaps with coefficients. First we try $$ (x^2 + x)^2 = x^4 + 2 x^2 + x^2 $$ Now we need $$ -4 x^2 - 4 x - 1 = -(2x+1)^2 $$ for $$ (x^2 + x)^2 - (2x+1)^2 = (x^2 + x + 2x + 1)(x^2 + x - 2x - 1) $$

Suppose I give a revised problem, use a different letter, factor $$ t^4 + 2 t^3 - 6 t^2 - 42 t - 63 $$ This does not factor over the rationals. However, it is $$ (t^2 + t)^2 - (7 t^2 + 42 t + 63) = (t^2 + t)^2 - 7( t^2 + 6 t + 9) = (t^2 + t)^2 - 7( t + 3)^2$$ or $$ \left( t^2 + t + \sqrt 7 (t+3) \right) \left( t^2 + t - \sqrt 7 (t+3) \right) $$ or $$ \left( t^2 + (1+\sqrt 7)t + 3 \sqrt 7 \right) \left( t^2 + (1-\sqrt 7)t - 3 \sqrt 7 \right) $$

Answer 2

When you attempt to complete the square of the first two terms and the opposite of the final two terms, you see that meet on the middle term. Therefore, you can express this as the difference of two squares.

$$a^4+2a^3 b-3a^2 b^2-4a b^3-b^4=(a^4+2a^3b+a^2b^2)-(b^4+2ab^3+4a^2b^2)\\=(a^2+ab)^2-(b^2+2ab)^2\\=\big((a^2+ab)+(b^2+2ab)\big)\big((a^2+ab)-(b^2+2ab)\big)\\=(a^2+3ab+b^2)(a^2-ab-b^2)$$

Answer 3

This solution is specific to this problem.

As others suggested, we look at $y=x^4+2x^3-3x^2-4x-1$.

We know $y'=4x^3+6x^2-6x-4$.

"Obviously" $x=1$ is a root of $y'=0$. Working out the rest, we find $y'=2(x-1)(2x+1)(x+2)$ and the zeros are $-2, -\frac 12,$ and $1$. We also notice that $-\frac 12$ is exactly halfway between the other two roots. This suggests that the polynomial may be symmetric about $x = -\frac 12$.

Letting $x = t-\frac 12$, we get

\begin{align} y &= t^4-\frac 92t^2 + \frac{1}{16} \\ &= \left(t^2 - \frac 94 \right)^2 - \frac{80}{16} \\ &= \left(t^2 - \frac 94 \right)^2 - 5 \\ &= \left(t^2 - \frac 94 - \sqrt 5 \right)\left(t^2 - \frac 94 + \sqrt 5 \right) \\ &\phantom{=} \text{Letting $t = x + \frac 12$, we get} \\ &= \left(x^2 + x - 2 - \sqrt 5 \right)\left(x^2 + x - 2 + \sqrt 5 \right) \\ \end{align}

Answer 4

$(a^2 - a b - b^2) (a^2 + 3 a b + b^2)$

A few comments. Note that every term is degree 4, so this is equivalent to factoring $$x^4+2x^3-3x^2-4x-1$$ You may recall the basic approach.

First look for integer solutions that divide the constant term (ie $\pm1$ in this case). Where the coefficient of the leading term is not 1, you look for factors $(ax-b)$ where $a$ divides the leading coefficient and $b$ divides the constant term,

If that succeeds, then it should be easy to divide out the linear factor. You then start again with the quotient.

If you fail to find a linear factor, then look for quadratic factors. Again you start by considering the leading term and the constant term. So in this case you are looking for a factorisation $$(x^2+hx+1)(x^2+kx-1)$$ and it is not hard to find it.