## Solving Polynomial Equations Up until now we have dealt with expressions, not equal to anything.  But, we are at a point, or should be, where we can begin explore polynomial equations, finding solutions, and learning about key features of polynomials.

This section is just an introduction that will be built upon when we more thoroughly explore quadratic equations (in the next section).  However, we will develop a deeper understanding of solving polynomial equations, and the key components to look for in doing so, in future sections.  But for now, this is an introduction to the basics.

Read through the notes, taking notes of your own.  Watch the videos, try the practice problems and be resourceful.  When you own your own learning you’re not bound by the imagination of your teacher!

One of the key differences between solving polynomial equations like 3x2 + 3x = 6, and most of the equations you’ve experienced until now is that the role of inverse operations is different.  Let me explain.

When you solve a single variable equation, like 3x + 6 = 4x + 1, your goal is to combine the terms with x together.  Because all of the terms with x are like terms, you can combine them with addition or subtraction.  However, when you’re solving something with the same variable, contained in unlike terms (different arrangement of exponents), you cannot combine them with addition and subtraction.

When we are solving equations we are looking for a value(s) that satisfies an equation, a value that makes it true.  We typically use inverse operations until we get x isolated, so we get x = #.  That’s when we’re done.  Things are different with many (but not all), polynomial equations.

A basic guiding principle is that you will use inverse operations to collect all terms on one side of the equation sign, so that the polynomial equals zero.  Then, you can try and factor.  Let’s see how that looks.

3x2 + 3x = 6

If we add 3 to both sides, we get:

3x2 + 3x – 6 = 0

There is no way to combine x2 and x, so that we can end up with x isolated.  We could isolate x or x2, but we’d get another x term on the other side of the equal sign.  So, what we’re going to do is factor.

Here we can take out a monomial, a constant term.

3x2 + 3x – 6 = 0

3(x2 + x – 2) = 0

Now, the remaining trinomial is factorable as well.

3(x + 2)(x – 1) = 0

Here we have three numbers, all multiplying, and the product is zero.  It certainly doesn’t look like three numbers, but it is.  We have a monomial and two binomials, all multiplying.

If we multiply three numbers together and the product is zero, then one of the numbers must be zero, right?  So, in 3(x + 2)(x – 1) = 0, either 3 = 0 (which is impossible, or x + 2 = 0, or x – 1 = 0. As it turns out, x is a variable, and can hold different values.  Our original equation will have two solutions, and each can be found by setting the factors equal to zero and solving them.

3(x + 2)(x – 1) = 0

Either x + 2 = 0    OR   x – 1 = 0.

For the first binomial, if x = -2, then we have a solution.  If x = 1, we have another solution.  The two solutions are -2 and 1.  Let’s check to see if they’re really solutions.  Remember, a solution is a value that makes the statement true.  So, let’s plug them into the original and see if they work.

3x2 + 3x = 6,

Check x = –2:  3(-2)2 + 3(-2) = 6

3(4) – 6 = 6

12 – 6 = 6 is true.

x = -2 is a solution.

Check x = 1:  3(1)2 + 3(1) = 6

3 + 3 = 6 is true.

Summary:  To solve a polynomial equation …

1. Use inverse operations to get it equal to zero.
1. Combine like terms, write it in descending order.
2. Factor the polynomial.
1. You have four methods to use
1. Factor a constant 