Solving equations that reduce to the form P·Q = 0
We use the fact that, if a product is equal to 0, then at least one of the factors must be 0. That is, if
then either
Note
This argument applies as well to a product of three or more terms; for instance, if
then
$P = 0$, $Q = 0$, or $R = 0$.
Examples
\t $4x^7-x^5 = 0$ \gap[10]
\\ \t $x^5(4x^2-1)=0$ \t \gap[10]
\\ \t Either $x^5=0$ or $(4x^2-1)=0.$ \t \gap[10]
\\ \t $x=0, x=-\frac{1}{2}$ #[or][o]# $x=\frac{1}{2}$ \t \gap[10]
\\ \\
\t $(2x+1)(x^2-4)-(x-3)(x^2-4) = 0$ \gap[10]
\\ \t $[(2x+1) - (x-3)](x^2-4) = 0$ \t \gap[10]
\\ \t $(x+4)(x^2-4) = 0$ \t \gap[10]
\\ \t $(x+4)(x-2)(x+2) = 0$ \t \gap[10]
\\ \t $x=-4, x=2$ #[or][o]# $x=-2$ \t \gap[10]
\\ \\
\t $x\sqrt{2x-1} = \sqrt{2x-1}$ \gap[10]
\\ \t $x\sqrt{2x-1} - \sqrt{2x-1} = 0$
\\ \t $\sqrt{2x-1}(x - 1) = 0$ \t \gap[10]