Recognizing Perfect Square Trinomials

Note that if a binomial of the form $a+b$ is squared, the result has the following form: $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2.$ So both the first and last term are squares, and the middle term has factors of $2, $ $a$, and $b,$ where the latter are the square roots of the first and last term respectively. 

For example, if the expression $2x+3$ were squared, we would obtain $(2x+3)(2x+3)=4x^2+12x+9.$ Note that the first term $4x^2$ is the square of $2x$ while the last term $9$ is the square of $3$, while the middle term is twice $2x\cdot3$.

It is important to be able to recognize such trinomials, so that they can the be factored as a perfect square. 

Factoring Perfect Square Trinomials

If you are attempting to to factor a trinomial and realize that it is a perfect square, the factoring becomes much easier to do. 

Example 1

Suppose you were trying to factor $x^2+8x+16.$ One can see that the first term is the square of $x$ while the last term is the square of $4$. Since the middle term is twice $4 \cdot x$, this must be a perfect square trinomial, and we can factor it as: 

$x^2+8x+16=(x+4)^2$

Example 2

Suppose you were trying to solve $9x^2+6x+1=0.$ You might try to factor the quadratic expression on the left-hand side of the equation. Since the first term is $3x$ squared, the last term is one squared, and the middle term is twice $3x\cdot 1$, this is a perfect square, and we can write:

$9x^2+6x+1=(3x+1)^2$ 

Thus the original equation has the form $(3x+1)^2=0$, so the only solution is when $3x+1=0$, which is when $3x=-1,$ or $x=-\frac{1}{3}. $ This quadratic equation has only that one solution.