# How does $(x+3)^2 - 2^2$ become $(x+1)(x+5)$?

I don't understand how $$(x+3)^2 - 2^2$$ can be transformed to equal $$(x+1)(x+5)$$. A short demonstration and/or reference to math rules would be very kind.

In general, $a^2-b^2=(a-b)(a+b)$. Let $a=x+3$ and $b=2$.

Just multiply each part out:

$(x + 3)^{2}$

$= (x + 3)(x + 3)$

$= x^{2} + 3x + 3x + 9$

$= x^{2} + 6x + 9$

Also, $2^{2} = 4$.

So, putting the two above together, we get:

$(x + 3)^{2} - 2^{2}$

$= x^{2} + 6x + 9 - 4$

$= x^{2} + 6x + 5$

And, if you understand how to factor, we get $x^{2} + 6x + 5 = (x + 5)(x + 1)$.

• Unnecessarily complex. Commented Feb 2, 2015 at 0:46
• @user314: We can verify many identities by a computation, and for this case the solution above shows that nicely. Commented Feb 2, 2015 at 0:53
• @AndréNicolas Of course, but the difference of squares factorization was way too apparent. Commented Feb 2, 2015 at 0:55

Yet another way of looking at it is to look at the roots.
$(x+3)^2-2^2=0$, so $(x+3)^2=2^2$. Taking square roots, we get $x+3=2$ or $x+3=-2$. Solving for x, we get $x=-1$ or $x=-5$. That means the quadratic is of the form $a(x+1)(x+5)$ where $a$ is the coefficient of your $x^2$ term, which in this case is 1.