What non-zero integer must be placed in the square so that the simplified product of these two binomials is a binomial: $(3x+1)(12x-\Box )$?
+592 If we need a binomial, then that means we need only 2 terms. One being \(36x^2\), and the other being an integer. This means we have to cancel out the terms with a variable of x.
Lets let the missing integer be y:
\((3x+1)(12x-y)=36x^2-3xy+12x-y\)
We just need to find an integer y such that \(-3xy+12x=0\) We can divide both sides by -3 and we see that y has to equal \(\boxed4\)
