Factor Polynomials

**Factor Polynomials**

To **factor an expression** means to treat the expression as a product, and write the factors which, when multiplied, yield that expression. Factoring is important because it can be used to solve equations and perform operations on fractions.

When factoring, there are three steps to keep in mind:

1. Always factor out the Greatest Common Factor (GCF).

2. Factor what is left.

3. If there are four terms, consider factoring by grouping.

**Example**

Factor the expression 6*x*^{4}*y* + 15*x*^{3}*y* − 9*x*^{2}*y*.

To follow step 1, look for the greatest common factor among the terms. We see that each coefficient (the number in front of each term) is evenly divisible by 3-- that is, 6, 15, and −9 all divide into 3 evenly. Let's factor it out and place it on the outside, and on the inside, the coefficients all get divided by 3:

3(2*x*^{4}*y* + 5*x*^{3}*y* − 3*x*^{2}*y*)

But we're not done. You can find the GCF among the variables by looking at each one's lowest exponent. In this case, the lowest exponent for *x* is 2 (seen in the third term), and the lowest exponent for *y* is 1 (technically, it's a three-way tie since all three terms have the same exponent for *y*, but it still counts). So factor out the *x*^{2}*y* along with the 3, and divide each term by that as well:

3*x*^{2}*y*(2*x*^{2} + 5*x* − 3)

Now we can move on to step 2. We need to factor 2*x*^{2} + 5*x* − 3. There are a number of strategies you can take to factor a three-term expression (*trinomial*) like this, so here's one possible solution:

Multiply the leading coefficient (2) by the constant term (−3):

2*x*^{2} + 5*x* − 3: (2)(−3) = −6

Now find two numbers that multiply to create this result, and at the same time, add together to make the middle coefficient (5). This may take a bit of guesswork:

−6 = −6 × 1; −6 + 1 = −5 **NO**

−6 = −2 × 3; −2 + 3 = 1 **NO**

−6 = −1 × 6; −1 + 6 = 5 **YES!**

Our numbers are −1 and 6. Now take the middle term (5*x*) and break it apart into −1*x* and 6*x*:

2*x*^{2} + 5*x* - 3 = 2*x*^{2} − 1*x* + 6*x* − 3

This is a popular method for factoring because it also involves step 3, using the 4-term expression for factoring by grouping. To factor by grouping, group the four terms into the **sum **of two 2-term expressions (*binomials*):

2*x*^{2} − 1*x* + 6*x* − 3 = (2*x*^{2} − 1*x*) + (6*x* − 3)

Now factor out the GCF from each binomial in the same way you did for step 1:

(2*x*^{2} − 1*x*) + (6*x* − 3) = *x*(2*x* − 1) + 3(2*x* − 1)

If you did this right, the two binomials you get should be exactly the same. All that's left to do now is factor *the binomial* out as if it were a GCF (which it is):

*x*(2*x* − 1) + 3(2*x* − 1) = (2*x* − 1)(*x* + 3)

Combine this with the GCF you took out of the original expression, and you have the fully factored result:

6*x*^{4}*y* + 15*x*^{3}*y* − 9*x*^{2}*y* = **3 x^{2}y(2x − 1)(x + 3)**

There are many types of factorable (or unfactorable) expressions. With practice, you'll get to know these different types and be able to perform these steps rather quickly. You may even find your own method of factoring that works even better for you. Try it!

**Practice**

Factor the following expressions completely.

1. 5x^{2} + 30*xy*

2. 12*p*^{3} − 8*p*^{2}

3. *x*^{2} + 7*x* + 10

4. 3*x*^{2} − 10*x* + 8

5. 2*t*^{2} + 10*t* − 72

6. 10*xy*^{3} − 55*xy*^{2} + 75*xy*

**Answers**

1. 5*x*(*x* + 6*y*)

2. 4*p*^{2}(3*p* − 2)

3. (*x* + 5)(*x* + 2)

4. (*x* − 2)(3*x* − 4)

5. 2(*t* + 9)(*t* − 4)

6. 5*xy*(2*y* − 5)(*y* − 3)