Solving a Quadratic Given in Factored Form

Think about why this works before you do it by habit. If two numbers multiply to give 0, one of them has to be 0 — there's no other way to get a product of zero from real numbers. So if (x-3)(x+5)=0, either (x-3)=0 or (x+5)=0. That single fact, the zero product property, is the whole engine behind this topic.

Mechanically, you split the factored equation into two tiny linear equations and solve each on its own. Take (2x-7)(x+4)=0. Set 2x-7=0 to get x=3.5, and set x+4=0 to get x=-4. Both values make the original product zero, so both are solutions — a quadratic in factored form hands you its roots almost for free, no formula needed. If the two factors happen to be identical, like (x-5)(x-5)=0, there's only one distinct solution, x=5, even though the quadratic still has degree two — that's called a repeated or double root.

This topic sits right before quadratics given in standard or vertex form, where the real work is getting to factored form first — by factoring, completing the square, or the quadratic formula. Once you're there, this same zero product move finishes the job. It also connects to graphing: the solutions you find here are exactly the x-intercepts of the parabola, since factored form is built to expose them.