$f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$. 
Let $f(x)$ and $g(x)$ be two monic cubic polynomials, and let $r$ be a real number. Two of the roots of $f(x)$ are $r+1$ and $r+7$. Two of the roots of $g(x)$ are $r + 3$ and $r + 9,$ and$$f(x) - g(x) = r$$for all real numbers $x.$ Find $r.$

So far, I have $$f(x)=(x-r-1)(x-r-7)(x-p)$$ and $$g(x)=(x-r-3)(x-r-9)(x-q).$$ From $f(x)-g(x)=r$, I know that their constant terms differ by $r$. I expanded the two functions but it was too complicated. I also plugged in $x=r+1,r+7,r+3,r+9$ into $f(x)-g(x)=r$, but it didn't do much.
Thanks in advance!!!!!

Your approach is fine. Note that\begin{multline}f(x)-g(x)=\\=(4-p+q)x^2+(2 r p+8p-12 q-2 q r-4 r-20)x-p r^2+q r^2-8 p r+12 q r-7 p+27 q.\end{multline}So, $4-p+q=0$; in other words, $p=q+4$. Replacing $p$ with $q+4$ in the coefficient of $x$ in $f(x)-g(x)$, we get that $4(3-q+r)=0$; in other words, $q=r+3$. And if replace $q$ with $r+3$ in the constant term of $f(x)-g(x)$, we get $32$. But we want this to be equal to $r$. Therefore, $r=32$.
