## Precalculus (6th Edition)

$(3m-2)^{2}$
You might recognize this a perfect square $(A-B)=A^{2}-2AB+B^{2}$ $A^{2}=9m^{2}=(3m)^{2}\Rightarrow A=3m$ $B^{2}=4=2^{2}\Rightarrow B=2$ $-2AB=-2(3m)(2)=-12m$ $9m^{2}-12m+4=(3m-2)^{2}$ If you don't recognize the perfect square, factor the trinomial as we have done so up to now: When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$ and, if we find them, we rewrite $bx$ and proceed to factor in groups. Here, factors of $(9)\times(4)=36$ that add to $-12$ are ....$-6$ and $-6 .$ $9m^{2}-12m+4=9m^{2}-6m-6m+4$ $=3m(3m-2)-2(3m-2)$ $=(3m-2)(3m-2)$ $=(3m-2)^{2}$