Factoring has been an important part in PreCalc 11 so far and will probably be in the future as well. Although we have had a lot of practice with them, there are some trinomials that are not as easy to factor as we are used to.

Example: 

(x-6)^2 + 7 (x-6) + 10

 

Step 1: Create a Variable in the Expression

The polynomials that we are used to factoring are the ones that have the pattern that follow ax^2 + bx + c. Even though we don’t have that pattern in this equation we can make it so that is does. We can do this by substituting something common. Something that’s common in this expression is the (x-6). We can make (x-6) equal any variable. In this case let’s make (x-6) = x. So wherever we see (x-6) we switch it for x

x^2 + 7x + 10

 

Now the expression follows the pattern that we are familiar with we can factor as usual.

Step 2: Factor

To factor we have to find a number that when added together equal seven and when multiplied equals 10. In this case the numbers are 5 and 2.

(x+2)(x+5)

 

Step 3: Substitute

We can’t forget about the term that we substituted. Since we made x = (x-6), everywhere we see x in the factored expression we have to switch it for (x-6)

$latex ((x-6) +2)((x-6)+5)

 

Step 4: Simplify

The only thing left to do is to simplify the expression.

(x-4)(x-1)