Intuition while factorizing an equation

Question

I have the following expression to factorize: $x^4 - 15x^2 + 36$

One way to solve it is:

$x^4 - 15x^2 + 36$

$x^4 - 12x^2 - 3x^2 + 36$

$x^2 (x^2 - 12) -3 (x^2 - 12)$

$(x^2 - 12) (x^2 -3)$

$(x-\sqrt{12})(x+\sqrt{12})(x-\sqrt{3})(x+\sqrt{3})$

$(x-2\sqrt{3})(x+2\sqrt{3})(x-\sqrt{3})(x+\sqrt{3})$

But, if I may start differently:

$x^4 - 15x^2 + 36$

$x^4 - 12x^2 - 3x^2 + 36$

$x^4 - 12x^2 + 36 - 3x^2$

$(x^2)^2 - 2(x^2)(6) + (6)^2 - 3x^2$

$(x^2-6)^2 - 3x^2$

$(x^2-6)^2 - (\sqrt{3}x)^2$

$(x^2-6-\sqrt{3}x)(x^2-6+\sqrt{3}x)$

Now I am sure these two results should be equivalent and can be converted to the other form (if my math is correct). But further steps become more complex to carry out. The first method gives the answer in the book. While 2nd will likely result in deduction of some marks by the grader. How can I intuitively start with the correct method?

Answer 1

A method is to use dummy variable substitution.

In quadratics this is simple as understanding how the roots influence the coeff of different $x$-terms.

If the roots are $k_1$ and $k_2$, then the general form of the quadratic can be expressed as $x^2 - (k_1 + k_2)x + k_1k_2$ or $x^2 - (\text{sum of roots})x + \text{product of roots}$. This can be obtained through the expansion of $(x - k_1)(x - k_2)$.

Now for example, $x^2 + \sqrt{3}x - 6$, you can try to brute factor $-6$ into irrational factors that summate to $\sqrt{3}$. Ie. $6 = -2\cdot3 = -2\sqrt{3}\cdot\sqrt{3}$ or $6 = 2\cdot-3 = 2\sqrt{3}\cdot-\sqrt{3}$. See that the latter sums up to what is required, $2\sqrt{3} - \sqrt{3} = \sqrt{3}$. Your latter method was not wrong after all. It was just incomplete.

For higher polynomials, one should break down into smaller polynomials until linear factors appear.

Consider, $$ x^4 - 15x^2 + 36 = (x^3 + ax^2 + bx + c)(x - d) \\\text{Notice the coeff of $x^3$ and $x$ is $0$} \\\implies (a - d) = 0 \iff a = d \\\text{ and } \\ c - bd = 0 \iff c = bd \iff c = ab \\\therefore (x^3 + ax^2 + bx + c)(x - d) = (x^3 + ax^2 + bx + ab)(x - a) \\=(x^2(x + a) + b(x + a))(x - a) = (x^2+ b)(x + a)(x - a) = (x^2 + b)(x^2 - a^2) $$

Observe the equation now resonates the simple quadratic factoring above. We continue to,

$$ a^2b = -36 = 12 \cdot -3 \text{ and } a^2 - b = 15 = 12 - (-3) \implies a \pm 2\sqrt{3}, \, b = -3 $$

After factoring $x^2 - 3$, we achieve all four linear factors.

Note a consequence of this method, is that it is time consuming, especially as the polynomial degree increases. As you have shown, the grouping technique is more nimble in finding roots for $x^4 - 15x^2 + 36$

Finally, I would not say there is intuition in a "correct" method. All methods that achieve correct factors would be "correct", just at the expense of time. As a matter of fact you could use the grouping technique on any polynomial. Consider this randomly generated polynomial, $x^3 - 2x^2 - 5x + 6$. Initially, it does not look like there are obvious grouping factors. However: $$ x^3 - 2x^2 - 5x + 6 \\=x^3 - 2x^2 - 5x + 6 \\=x^3 + x^2 - 2x - 3x^2 - 3x + 6 \\=x(x^2 + x - 2) - 3(x^2 + x - 2) \\=(x - 3)(x^2 + x - 2) \\=(x - 3)(x^2 - x + 2x - 2) \\=(x - 3)(x(x - 1) + 2(x - 1)) \\=(x - 3)(x + 2)(x - 1) $$ Of course, it could have consumed more time as this grouping technique is not suited to the original polynomial but succeeds with some algebraic manipulation.