# Factoring Expressions

## Examples with solutions

**Example 1: **

Factor: 45x^{ 5}y^{ 2} – 125xy^{ 4}.

**solution:**

This expression is the difference of two terms, but neither appear to be perfect squares. However, we should not abandon this problem immediately, because we haven’t really applied the general strategy.

So, first check for common monomial factors. You can see that
both terms share factors of 5, x, and y^{ 2}. Thus

45x^{ 5}y^{ 2} – 125xy^{ 4} = 5xy^{
2}(9x^{ 4} – 25y^{ 2})

Now, having completed step (i) of the general strategy, we can check the remaining part of the original expression,

9x^{ 4} – 25y^{ 2},

against the special patterns listed for step (ii). But now we really do have a difference of two perfect squares:

9x^{ 4} – 25y^{ 2} = (3x^{ 2})^{
2} – (5y)^{ 2}

and so

9x^{ 4} – 25y^{ 2} = (3x^{ 2} +
5y)(3x^{ 2} – 5y)

Thus, our original expression can now be written as

45x^{ 5}y^{ 2} – 125xy^{ 4} = 5xy^{
2}(9x^{ 4} – 25y^{ 2})

= 5xy^{ 2}(3x^{ 2} + 5y)(3x^{ 2}
– 5y)

Before concluding that our work is done, we need to look
briefly at each non-monomial factor: (3x^{ 2} + 5y) and
(3x^{ 2} – 5y). Neither contain monomial factors,
since those have already been isolated for the overall
expression. Then

- (3x
^{ 2}+ 5y) is a sum of two terms, and we have no formula for this pattern

and

- (3x
^{ 2}- 5y) is a difference of two terms, but neither one is a perfect square

Thus, no further factorization is possible, and so our final answer here is

45x^{ 5}y^{ 2} – 125xy^{ 4} = 5xy^{
2}(3x^{ 2} + 5y)(3x^{ 2} – 5y)

(To verify that this is a correct answer, you should probably take a minute to multiply out the righthand side to remove all brackets. You should find that, indeed, the original expression on the lefthand side is regenerated, confirming that the two expressions really are mathematically equivalent.)

**Example 22:**

Factor: 4x^{ 3}y + 8x^{ 2}y + 28xy.

**solution:**

All three terms of this expression share monomial factors of 4, x, and y. Thus

4x^{ 3}y + 8x^{ 2}y + 28xy = 4xy(x^{ 2}
+ 2x + 7)

What is left after removing the common monomial factors is the trinomial in x:

x^{ 2} + 2x + 7

When we try to find whole numbers, a and b, that enable us to write this trinomial as a product of the form (x + a)(x + b), we find that there are no such values. (We’d need ab = 7, and a + b = 2, and it is easy to check that none of the pairs of whole numbers that multiply to give 7 also add up to give 2.) This means that the trinomial cannot be factored, and so the most complete factorization of the original expression is the factorization already stated above:

.4x^{ 3}y + 8x^{ 2}y + 28xy = 4xy(x^{ 2}
+ 2x + 7)

Do you need help with solving your algebra homework? DoMyHomeworks service is there to complete it for you.