How do you factor $(10x+24)^2-x^4$?

Question

I tried expanding then decomposition but couldn't find a common factor between two terms

Answer 1

Hint: $$a^2-b^2=(a+b)(a-b)$$ Work out what your $a$ and $b$ should be.

Answer 2

Using the last hint:

$(10x+24)^2-x^4=((10x+24)+x^2)((10x+24)-x^2)$

Answer 3

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!!!