Solving Polynomial Equations by Factoring

Solve Quadratic Equations by the Zero-Factor Property

A quadratic equation is an equation that can be written in the form

ax2 + bx + c = 0

where a, b, and c are real numbers, with a ≠ 0.

ax2 + bx + c = 0 is called the standard form of a quadratic equation. Sometimes we may have to rearrange the terms of the equation to put it in standard form. The only requirement here is that we have an x2 in the equation. We guarantee that this term will be present in the equation by requiring a ≠ 0.

Note however, that it is okay if b and/or c is zero.

The following equations are examples of quadratic equations.

1.2k2 + 4k + 1 = 0a = 2, b = 4, c = 1

2.x2 – 12x = 0a = 1, b = – 12, c = 0

3.t2 = 81a = 1, b = 0, c = – 81

4.(y + 1)(y – 7) = 1a = 1, b = – 6, c = – 8 (Hint: Expand and collect terms).

The Zero-Factor Property

Solving by Factoring As the heading suggests, we will be solving quadratic equations here by factoring them. To do this we will need the following fact.

If ab = 0 then either a = 0 and/or b = 0.

This fact is called the zero factor property or zero factor principle. All that the property says is that if a product of two terms is zero then at least one of the terms had to be zero to start off with. Notice that this property will ONLY work if the product is equal to zero. Consider the following product.

ab = 6

In this case there is no reason to believe that either a or b will be 6.

We could have a = 2 and b = 3 for instance. So, do not misuse this fact!

To solve a quadratic equation by factoring we first must move all the terms over to one side of the equation.

Doing this serves two purposes. First, it puts the quadratics into a form that can be factored. Secondly, and probably more importantly, in order to use the zero factor property we MUST have a zero on one side of the equation.

If we don’t have a zero on one side of the equation we won’t be able to use the zero factor property.

Example 1: Solve the following equation by factoring. x2x = 12

Solution.

First, get everything on one side of the equation and then factor.

x2x – 12 = 0 (x – 4)(x + 3) = 0 Now at this point we’ve got a product of two terms that is equal to zero. This means that at least one of the following must be true.

x – 4= 0ORx + 3 = 0

x = 4ORx = – 3

Note that each if these is a linear equation that is easy enough to solve. What this tells us is that we have two solutions to the equation, x = 4 and x = – 3. As with linear equations we can always check our solutions by plugging the solution back into the equation. We will check x = – 3 and leave the other to you to check.

(– 3)2 – (– 3) = 12 ?

9 + 3 = 12 ?

12 = 12 OK

So, this was in fact a solution.

Example 2: Solve the following equation by factoring.

4m2 – 1 = 0

Solution.

Factor the equation. 4m2 – 1 = 0

(2m – 1)(2m + 1) = 0

Now apply the zero factor property. The zero factor property tells us that,

2m – 1= 0OR2m + 1 = 0

2m = 1 OR2m = – 1

m = m = -\frac{1}{2} ORm = – m = -\frac{1}{2}

Again, we will typically solve these in our head, but we needed to do at least one in complete detail. So we have two solutions to the equation.

m = m = -\frac{1}{2} ANDm = – m = -\frac{1}{2}

Example 3: Solve the following equation by factoring.

5x³ – 5x2 – 10x = 0

Solution.

The first thing to do is factor this equation as much as possible. In this case that means factoring out the greatest common factor first. Here is the factored form of this equation.

5x(x2x – 2) = 0

5x(x – 2)(x + 1) = 0

Now, the zero factor property will still hold here. In this case we have a product of three terms that is zero. The only way this product can be zero is if one of the terms is zero. This means that,

5x = 0 —› x = 0

x – 2 = 0 —› x = 2

x + 1 = 0 —› x = – 1

So, we have three solutions to this equation. So, provided we can factor a polynomial we can always use this as a solution technique. The problem is, of course, that it is sometimes not easy to do the factoring.

Solve 2x 2 +3x=2

Which of the following equations has both 1 and -3 as solutions?