ex1.5_1

$x$ $\quad (2x+1)(x-5)=0$

The expression on the left is completely factored, and there is a zero on the right. We can solve by setting each factor equal to zero.

(2x+1)(x-5)=0\\[0.25 in] 2x+1=0 \quad \mbox{or} \quad x-5=0\\[0.25 in] x=-\frac{1}{2} \quad \mbox{or} \quad x=5