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?
 A: Hint / Nudge:
It depends on what you mean by "best," but there is a special formula for factoring the difference of two perfect squares:
$$a^2 - b^2 = (a-b)(a+b)$$
In your case, notice:
$$25x^2 - 121 = (5x)^2 - (11)^2$$
From here, you can factor the expression accordingly.

Similar Question:

Let's factor the following: $$9x^4 - 16$$

Notice:
$$9x^4 - 16 = (3x^2)^2 - (4)^2$$
This is the difference of two perfect squares! In our above formula, take $a = 3x^2, b = 4$ and we find
$$9x^4 - 16 = (3x^2 - 4)(3x^2 + 4)$$
Note, too, that if you let $3 = (\sqrt 3)^2$, we can factor the left one even more if you wanted. This task is an exercise left to the reader, though. ;) 
(And of course, generally, when you're expected to factor, often you want to avoid having square roots since that introduces various problems. But it's a nice exercise to show you it can be done.)

Alternate Method:
If you want to use more "standard" methods of factoring... well, let's recall what we call "standard":


*

*We begin with a trinomial, $ax^2 + bx + c$.

*We find factors of $a \cdot c$ that add up to $b$. Let those factors be $p, q$ for sake of argument.

*Then $ax^2 + bx + c = (x + p/a)(x + q/a)$.

*Next if the fractions $p/a$ and $q/a$ have a denominator even after simplifying, you'll want to "move" that denominator next to $x$. For example, $(x + 2/5)$ turns into $(5x+2)$.


This is the method we're all familiar with. So suppose, for example, in our case, we're looking at $25x^2 - 121$ as given. We notice:
$$25x^2 - 121 = 25x^2 + 0x - 121$$
In this light we can factor this expression using the old method. (This exercise is left to the reader. :p)
Personally, the difference of squares method is faster and simpler, but this should work too.
A: 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 $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.
