## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\left( x-\dfrac{1}{5}\right)^2$
The 2 numbers whose product is $ac= 1\left( \dfrac{1}{25} \right)=\dfrac{1}{25}$ and whose sum is $b= -\dfrac{2}{5}$ are $\left\{ -\dfrac{1}{5}, -\dfrac{1}{5} \right\}.$ Using these two numbers to decompose the middle term of the given trinomial, $x^2-\dfrac{2}{5}x+\dfrac{1}{25} ,$ then the factored form is \begin{array}{l} x^2-\dfrac{1}{5}x-\dfrac{1}{5}x+\dfrac{1}{25} \\\\= \left( x^2-\dfrac{1}{5}x \right) -\left( \dfrac{1}{5}x-\dfrac{1}{25}\right) \\\\= x\left( x-\dfrac{1}{5} \right) -\dfrac{1}{5}\left( x-\dfrac{1}{5}\right) \\\\= \left( x-\dfrac{1}{5}\right)\left( x-\dfrac{1}{5}\right) \\\\= \left( x-\dfrac{1}{5}\right)^2 .\end{array}