Intermediate Algebra (12th Edition)

$(3m-5n)^2$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $9m^2-30mn+25n^2 ,$ use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ In the trinomial expression above, $a= 9 ,b= -30 ,\text{ and } c= 25 .$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac= 9(25)=225$ and whose sum is $b$ are $\left\{ -15,-15 \right\}.$ Using these two numbers to decompose the middle term results to \begin{array}{l}\require{cancel} 9m^2-15mn-15mn+25n^2 .\end{array} Using factoring by grouping, the expression above is equivalent to \begin{array}{l}\require{cancel} (9m^2-15mn)-(15mn-25n^2) \\\\= 3m(3m-5n)-5n(3m-5n) \\\\= (3m-5n)(3m-5n) \\\\= (3m-5n)^2 .\end{array}