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It's easy enough to show that $(a+b)(a-b) = a^2 - b^2$, you just simply multiply it out:

$$(a + b)(a - b)$$ $$= a^2 -ab + ba - b^2$$ $$= a^2 - b^2$$

However, is there any way using basic algebra that you can go from $a^2 - b^2$ to $(a + b)(a - b)$? Or is this something that you're just supposed to know?

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  • $\begingroup$ add $0=ab-ba$, namely reverse the equality you had... $\endgroup$
    – TTY
    Commented Mar 13, 2014 at 7:54
  • $\begingroup$ The subject of mathematics consists of many things that one learns by being taught, such as factoring the difference of two squares. This is more or less the meaning of the root word "mathesis"... $\endgroup$
    – hardmath
    Commented Mar 13, 2014 at 10:26

3 Answers 3

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The steps in the first argument you gave are reversible. You can add $-ab+ba$ to the left side, because it equals $0$, and then do your factoring.

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  • $\begingroup$ I guess what I'm trying to ask is, how does someone just 'know' that you have to add $-ab+ba$ to the equation? It just seems so arbitrary. Like why those 2 values and not, say, $-2ab+ba$? $\endgroup$
    – Daniel T.
    Commented Mar 13, 2014 at 8:00
  • $\begingroup$ It helps to have seen it before. Otherwise, it would seem very arbitrary. What makes it not arbitrary is that it works. $\endgroup$ Commented Mar 13, 2014 at 8:00
  • $\begingroup$ @DanielT.: Keep in mind we can't add just any expression, say $-2ab + ba$; it must be equal to zero in order for the value to stay the same. $\endgroup$
    – hardmath
    Commented Mar 13, 2014 at 10:23
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Let $$P(x)=x^2-b^2$$ then we see that $P(b)=P(-b)=0$ then $b$ and $-b$ are two roots of $P$ which's with degree 2 then we factorize $$P(x)=x^2-b^2=(x-b)(x+b)$$ Now take $x=a$.

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Make a square of length 'a' and draw a square of length 'b' inside the first square so that the edges overlap. Name the first square as ABCD and the second one as AEFG with AE on AB and AG on AD. Produce EF to meet CD at H and GF to meet BC at I. Then $$a^2-b^2=[ABCD]-[AEFG]$$ $$=[EFIB]+[DHFG]+[FICH]$$ $$=2b(a-b)+(a-b)^2$$ $$=(a-b)(a-b+2b)=(a-b)(a+b)$$

where [] denotes the area of the figure.

You can also rotate and translate FGDH so that FH coincides with CH. Then EBGD becomes a rectangle. $$a^2-b^2=[EBGD]=EB*ED=(a-b)(a+b)$$

Other algebraic identities can also be found using this method like for $(a+b)^2,(a-b)^2,etc$.

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