<s>[INST] What is the best way to factor $25x^2-121$? What is the best way to factor $25x^2-121$ ?
The methods I have been taught doesn't work for this problem. Is there an another way to solve this problem? If so, how?
 [/INST] The hard way:
We want to factor, IF it can be done, as
$25x^2 -121 = (ax + b)(cx + d) = acx^2 + (bc+ad)x + bd$ where
$ac = 25;  bd= -121; bc+ad= 0$. 
IF this can be factored to $ac =25$ means $a,c$ can be factored as $1,25$ or $5,5$.
And $bd =121$ means $b,d$ can be factored to $\pm 1;\mp 121$ or $\pm 11, \mp 11$.
But we have $ad = -bc$  of the options, if $a=25$ then $b=1$ and we can't have $25d = -c$ as $c$ can't have anything to do with $5$. 
The same issues occur if $b = 121$. 
To get $ad = -bc$ it helps if $|a| = |c| = 5$ and $|b|=|d| = 11$.  Then everything is nicely balanced and should cancel out.
And indeed:  $(5x + 11)(5x - 11) = 5x\times 5x + 11\times 5x + 5x\times(-11) + 11\times(-11) = 5x^2 - 121$.
The easy way:
Remember the "difference of squares" rule:  $a^2 - b^2 = (a+b)(a-b)$.
So $25x^2 = (5x)^2$ and $121 = 11^2$ so $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
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