$g(x)$ has zeros at $x=r_1+5,~x=r_2+5,~and~x=r_3+5$
Work Step by Step
We have that: $f(r_1)=f(r_2)=f(r_3)=0$ $g(r_1+5)=f(r_1+5-5)=f(r_1)=0$ $g(r_2+5)=f(r_2+5-5)=f(r_2)=0$ $g(r_3+5)=f(r_3+5-5)=f(r_3)=0$
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