Applying the Perfect Square Identity

The perfect square forms $(a+b)^2$ and $(a-b)^2$ come up a lot in algebra. We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often.

Expand $(a+b)^2$.

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We have

$$\begin{aligned} (a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ & = a^2 + ab + ba +b^2 \\ &= a^2 + 2ab +b^2.\ _\square \end{aligned}$$

Expand $(a-b)^2$.

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We have

$$\begin{aligned} (a-b)^2 &= (a-b)(a-b) \\ &= a(a-b) - b(a-b) \\ & = a^2 - ab - ba +b^2 \\ &= a^2 - 2ab +b^2.\ _\square \end{aligned}$$

Expand $(x+2)^2$.

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We have

$$\begin{aligned} (x+2)^2 &= (x+2)(x+2) \\ &= x(x+2) + 2(x+2) \\ & = x^2 + 2x + 2x +4 \\ &= x^2 + 4x +4.\ _\square \end{aligned}$$

For these problems, you will need to recognize the perfect square form, in order to quickly solve it.

Evaluate $73 ^2 + 2 \times 27 \times 73 + 27 ^2$.

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Observe that with $a = 73, b = 27$, we obtain $a^2 + 2 \times b \times a + b^2 = ( a + b) ^2 = 100 ^2 = 10000$. $_ \square$

Factorize $n^4 + 4$.

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This isn't immediately a perfect square as yet. If we attempt to complete the square, we see that we need the term $4 n^2$. So, let's add and subtract this to obtain

$$n^4 + 4n^2 + 4 - 4n^2 = ( n^2 + 2) ^2 - 4 n^ 2.$$

We now apply the difference of two squares identity, to conclude that

$$( n^2 + 2) ^2 - 4 n^ 2 = ( n^2 + 2)^2 - (2n) ^2 = ( n^2 + 2 + 2n) ( n^2 + 2 - 2n ).\ _\square$$

The above factorization is known as the Sophie-Germain identity:

$$n^4 + 4 = ( n^2 + 2n + 2 ) ( n^2 - 2n + 2 ).$$