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Solving Polynomial Equations

There's a very simple principle involved in solving polynomial equations: If two (or more) numbers multiplied together equal zero then at least one of the numbers has to be zero. Don't take my word for this. Try to find two non-zero numbers that, when you multiply them together, give you zero. I'm pretty sure you won't be able to do it. The magic part, for us, is that this works for products that have variables in them too. If I tell you that xy = 0 then at least one of x and y has to be zero. Here's how we use that to solve a polynomial equation:

Warning - This Only Works with 0

Keep in mind that you can't do this with just any number on the right hand side of the equation. It only works when the equation equals zero. You can't, for example, take something like x(x + 1) = 2 and solve it by setting x = 2 and x + 1= 2.

  1. Set the equation equal to zero if it isn't already.
  2. Factor the polynomial.
  3. Set each of the factors from step two equal to zero.
  4. The solution to the smaller equations you came up with in step three are the solution to the original equation.

Example 1

Solve the equation x2 + 3x - 4 = 0.

This equation is already equal to zero so we can skip directly to step two. The first thing we need to do here is factor the polynomial.

(x + 4)(x - 1) = 0

Now we need to take each of the two factors, x + 4 and x - 1 and make two new equations by setting each of them equal to zero.

x + 4 = 0x - 1 = 0

Solving the first equation gives us x = -4; solving the second one gives us x = 1. Those two numbers, x = -4, 1 are the solutions to our original equation.

Example 3

Solve the equation 6x2 - 19x + 10 = -5.

In this equation, it isn't equal to zero so the first thing we need to do is make it equal to zero. We can do that by adding 5 to both sides.

6x2 - 19x + 10 = -5
6x2 - 19x + 15 = 0

Now we can factor the polynomial to get:

(2x - 3)(3x - 5)

Finally, we need to take each of the two factors, 2x - 3 and 3x - 5, and make two new equations by setting each of them equal to zero.

2x - 3 = 03x - 5 = 0

Solving the first equation gives us x = 3/2; solving the second one gives us x = 5/3. Those two numbers, x = 3/2, 5/3 are the solutions to our original equation.

Example 2

Solve the equation x2 - 2x + 1 = 0.

This equation is also already equal to zero so we can skip directly to step two again. The polynomial on the left hand side factors to:

(x - 1)(x - 1) = 0

Now we need to take each of the two factors, x - 1 and x - 1 and make two new equations by setting each of them equal to zero.

x - 1 = 0x - 1 = 0

I'm sure by now you've noticed that the two factors and therefore the two equations are the same. Don't let that bother you - sometimes a quadratic equation is going to have only one answer instead of two. In this case, the answer would be only x = 1.

Example 4

Solve the equation x2 + 2x + 1 = 0.

This equation is also already equal to zero so we can skip directly to step two again. At this point, we're going to run into some trouble. If you work on x2 + 2x + 1 using the techniques from our discussion of factoring, you'll come to the conclusion that it can't be factored. Sometimes that's going to happen. When it does, your conclusion should be that there aren't any real numbers that are solutions to the equation.

Quick Tip - Looking Ahead

The equation in Example 4 only doesn't have solutions if we limit our answers to real numbers. If we expand our thinking to complex numbers then it will have solutions but you won't be able to find them by factoring. For that, you need the techniques that we'll discuss in the sections on the Quadratic Formula.

Dynamic Practice - Solving by Factoring