What is the best way to factor $25x^2-121$?

Question

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?

Answer 1

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.

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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.)

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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.

Answer 2

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 $25x^2 -121 = (5x)^2 - (11)^2 = ((5x) + (11))((5x) - (11)) = (5x +11)(5x-11)$.