Factoring Expressions

In order to understand the following lesson, you should already know:

(i.) a general method for removing monomial factors from an expression

(ii.) two methods for factoring expressions with very special (but quite commonly occurring) forms:

• trinomials (ax ² + bx + c)

• difference of two perfect squares (u ² – v ² )

All we want to point out here is that the overall strategy for factoring any algebraic expression is to apply the methods listed above in order. So,

(i) first remove all monomial factors from the expression

then

(ii) check whatever is left against any of the special patterns that you know

By following this strategy, the more difficult, tedious, and intuitive analysis of the second type in the list above will always be done on the simplest possible expressions.

We’ve already alluded to this strategy in the previous documents on factoring, and shown one or two brief examples. Here is a few more short examples to help you understand the overall strategy better.

Example 1:

Factor completely: 3x ⁴ y + 6x ³ y – 45x ² y.

solution:

We see that 3, x ² , and y, are each monomial factors of all three terms. So, step (i) in the general strategy leads to

3x ⁴ y + 6x ³ y – 45x ² y = 3x ² y(x ² + 2x – 15)

Now, we need to apply step (ii) of the general strategy to the expression, x ² + 2x – 15, remaining after we have accounted for the common monomial factors. This is a trinomial in x, which it may be possible to rewrite in the form

x ² + 2x – 15 = (x + a)(x + b)

if we can find two whole numbers, a and b, such that

ab = -15 and a + b = -2.

You can easily verify that a = 5 and b = -3 works. Thus

x ² + 2x – 15 = (x + 5)(x - 3)

Thus, for the original expression we now have

3x ⁴y + 6x ³y – 45x ²y = 3x ²y(x + 5)(x – 3)

which is factored as completely as is possible. We know that no further factorization is possible, because the factors here which are not monomial factors contain no products at all.

Notice that when it came time to check if any of our special pattern formulas might apply, we needed to check them against the quite simple expression, x ² + 2x – 15, rather than the much more complicated original expression. To be sure, (x + 5) and (x – 3) are factors of the original expression, but its complexity (due to the presence of common monomial factors in each term) would make it very difficult to recognize these binomial factors in that expression.