Let \(a\) and \(b\) be real numbers such that \(a\cdot b=0\). In order for this equation to be true, either \(a\) or \(b\) must be 0.
We can use this fact to solve factored equations such as \((x-1)(x+2)=0\), knowing that either \((x-1)=0\) or \((x+2)=0\). Each of these equations implies that \(x=1\) and \(x=-2\) respectively, simply by solving for \(x\) in each equation. This works because \(x\) represents an unknown number, so therefore, \((x-1)\) and \((x+2)\) are unknown real numbers, just like the \(a\) and \(b\) in the first paragraph.
In fact, any time you have a product of two expressions such as \((blah)(yadayada)=0\), the zero factor property guarantees that either \((blah)\) or \((yadayada)\) is \(0\) allowing you to solve two separate equations.
CAVEAT! This method only works when one side of the equation is zero and the other side is completely factored! If you have something like \((x-1)(x+3)=12\), this does NOT imply that one of \((x-1)=12\) or \((x+3)=12\). To see this, plug in \(x=3\) into \((x-1)\) and \((x+3)\) to get \(2\) and \(6\) respectively for each; neither of which is \(12\).
More often than not, you will be using this zero factor property to solve quadratic equations. Specifically, this property will come into play once you’ve gotten everything in a quadratic equation to one side, then factored the nonzero side completely.
For example, suppose you want to solve the equation \(x^2-5x+5=-1\). Use the following steps:
NOTE: Not every quadratic equation can be solved in this way. Only the ones that can be factored once you get everything over to one side can be solved this way. Otherwise, you will have to use the methods found in the next couple lessons.