<s>[INST] How do you factor $(10x+24)^2-x^4$? I tried expanding then decomposition but couldn't find a common factor between two terms
 [/INST] Whoa there, Ladies and Gentlemen!  We ain't done yet!  As pointed out by yanbo and Hugus in their answers, a good first step is to observe that the identitiy
$a^2 - b^2 = (a + b)(a - b) \tag{1}$
may be directly applied to
$(10x+24)^2-x^4 \tag{2}$
and we obtain
$(10x+24)^2-x^4 = (10x + 24 + x^2)(10x + 24 - x^2); \tag{3}$
but $10x + 24 + x^2$ and $10x + 24 - x^2$ can be factored as well!  We have
$10x + 24 + x^2 = (x + 4)(x + 6), \tag{4}$
and 
$10x + 24 - x^2 = -(x + 2)(x - 12) = (x + 2)(12 - x). \tag{5}$
Apparently $(10x+24)^2-x^4$ may be completely reduced over $\Bbb Z[x]$ to linear factors:
$(10x+24)^2-x^4 =  -(x + 4)(x + 6)(x + 2)(x - 12), \tag{6}$
and its roots are $-6, -4, -2 \, \text{and} \, 12$.  Cute, the way $10x + 24$ is always a perfect square for any of these integers; but, then again, that's factoring in $\Bbb Z[x]$ for you.
Hope this helps!  Cheerio, 
and as always,
Fiat Lux!!!
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