## Intermediate Algebra (12th Edition)

$k^2-\dfrac{10}{7}kp+\dfrac{25}{49}p^2$
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $\left( k-\dfrac{5}{7}p \right)^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (k)^2+2(k)\left(-\dfrac{5}{7}p\right)+\left(-\dfrac{5}{7}p\right)^2 \\\\= k^2-\dfrac{10}{7}kp+\dfrac{25}{49}p^2 .\end{array}