Solving Quadratics With Factoring


The reason you’ve learned about factoring trinomials at this point in your Algebra 1 career is that it can be used to solve a lot of quadratic equations algebraically. We’re going to build on our ability to factor with a little bit of mathematical logic to learn how to solve more complicated equations in just a few easy steps.

The first step is going to be to get the equation in the ax² + bx + c = 0 format. This is necessary because our logic for solving equations is going to rely on having everything set to equal zero a few steps down. Simply move everything to one side of the equals sign and you’re in good shape.

The next step is going to be to factor the polynomial you’re left with. If you’re got 6x² + 11x – 10 = 0, and you figure out that when you factor 6x² + 11x – 10 you get (2x + 5)(3x – 2), then your new equation is (2x + 5)(3x – 2) = 0.

Now that we have the equation factored, consider the following. If a set of values are multiplied together to equal zero, what does that tell us about the values? For example, if we have ab = 0, what does that tell us about the variables a and b? It tells us that either a, b, or both are equal to zero. So we know that a = 0 or b = 0, or both.

Going back to our example, we know after factoring that (2x + 5)(3x – 2) = 0. So what does that tell us about 2x + 5 and 3x – 2? It tells us that 2x + 5 = 0 or 3x – 2 = 0, or both. Now we have some simple equations we have solved before, and we get the following:

2x + 5 = 0 
2x = -5 
x = \frac{-5}{2} 

3x - 2 = 0 
3x = 2 
x = \frac{2}{3} 

If you remember back to our initial discussion of solving quadratics, you’ll recall that we said some quadratics will have two possible answers. Now that we have two possible answers, let’s check them both by plugging them back into the original equation to see if we get something that makes sense. Let’s start with x = -5/2:

6x^2 + 11x - 10 = 0 
6(\frac{-5}{2})^2 + 11(\frac{-5}{2}) - 10 = 0 
6(\frac{25}{4}) + 11(\frac{-5}{2}) - 10 = 0 
\frac{150}{4} - \frac{55}{2} - 10 = 0 
\frac{75}{2} - \frac{55}{2} - \frac{20}{2} = 0 
\frac{75 - 55 - 20}{2} = 0 
0 = 0 

We see that x = -5/2 checks out. So now let’s try x = 2/3:

6x^2 + 11x - 10 = 0 
6(\frac{2}{3})^2 + 11(\frac{2}{3}) - 10 = 0 
6(\frac{4}{9}) + \frac{22}{3} - 10 = 0 
\frac{24}{9} + \frac{22}{3} - 10 = 0 
\frac{24}{9} + \frac{66}{9} - \frac{90}{9} = 0 
\frac{24 + 66 - 90}{9} = 0 
0 = 0 

And we also see that x = 2/3 checks out. Both of these are solutions to the original equation, and we figured this out by adding one step on to factoring.