Question

What values would make a ploynimal equal to zero if the factors are (x+3) and (x+12)

Answer 1

`\bf (x+3)(x+12) =0 \\ \quad \\ \begin{cases} (x+3)=\cfrac{0}{(x+12)}\to x+3=0\to &x=-3 \\ \quad \\ (x+12)=\cfrac{0}{(x+3)}\to x+12=0\to &x=-12 \end{cases}`

Answer 2

The values 35 and 25 would make the answer to this 0