## Precalculus (6th Edition)

$c=9$
RECALL: There are two forms of perfect square trinomials: (1) $m^2+2mn+n^2$, which is the square of $(m+n)^2$; and (2) $m^2-2mn+n^2$, which is the square of $(m-n)^2$ The given trinomial has: $m^2 = 100r^2=(10r)^2$, which means that $m=10r$ The middle term of the trinomial is negative therefore the perfect square trinomial will be in the same form as in (2) above. Thus, $-2mn=-60r$ Since $m=10r$, substitute $m$ with $10r$ to obtain: $-2mn=-60r \\-2(10r)(n) = -60r \\-20r(n) = -60r$ Divide both sides by $-20r$ to obtain: $\dfrac{-20r(n)}{-20r} = \dfrac{-60r}{-20r} \\n=3$ The given trinomial is in the form $m^2-2mn+n^2$. Square $n$ to obtain: $n^2 = 3^2 = 9$ Therefore, the perfect square trinomial is $100r^2-60r+9$. Hence, $c=9$