#### Factoring Polynomials

*What is Factoring?*

Recall the distributive law (BR-9) from the Basic Rules of Algebra section. This is generally called "expanding" while doing this rule in reverse is called "factoring". The expressions to either expand or factor are usually more complex than a(b + c) = ab + ac, but the same procedure is followed (recall the "FOIL" method discussed in the Basic Rules of Algebra section).

Essentially, *factoring* is rewriting an expression as a product of 2 or more expressions. [Recall back to elementary mathematics where factoring a number, such as 15, would be writing it as the multiple of two other numbers, such as 15 = (5)(3) where both 5 and 3 are factors of 15.] While a *polynomial* is an expression involving powers of x (or any variable) that involves only addititon and multiplication as the types of arithmetic operations used. Each term in a polynomial can be written as ax^{j} where a is a real number and j is a non-negative integer. There is more information about this in the Polynomials and Roots section.

*Common Factors*

One of the most basic ways to factor an expression is to "take out a common factor". If every term in an expression has several factors, and if every term has at least one factor that is the same, then that factor is called a *common factor*. If this is the case, then the common factor can be "taken out" of every term and multiplied by the whole remaining expression.

To see this clearly, lets look at an example: 2x^{2} + 8x

The first term has factors of 2, x, x and the second term has factors of 2, 4, x. It is apparent that 2 and x are both common factors so we call 2x the common factor. In this case, 2x^{2} + 8x = 2x(x + 4). Click here to see more examples.

*Decomposition Method*

Many people may learn how to factor by using trial and error, but that is sometimes difficult to grasp and is hard to explain. The *decomposition method* is a systmatic method used for factoring quadratic equations. It is best to look at what happens when expanding an expression, in order to see what needs to be done in reverse to factor it:

Expanding: (x + 5)(x + 1) = x(x + 1) + 5(x + 1) = xIn order to do this, you must have the x term split into two seperate terms. Obviously there are many ways to do this but only one ways works. In the example above, 6x was broken into 5x + 1x but if it was broken into 2x + 4x, we would get a different answer (that would be incorrect). [Try it and prove to yourself that it will not work.] The rules to split up the x term correctly are as follows:^{2}+ x + 5x + 5 = x^{2}+ 6x + 5 Factoring, starting from x^{2}+ x + 5x + 5: = x(x + 1) + 5(x + 1) = (x + 1)(x + 5)

Given a general form of a quadratic expression, axOnce you have split the x term, you continue factoring by taking out a common factor from the first two terms, and then a common factor from the last two terms. Your expression should now look like: x(x + n) + m(x + p) where n, m, and p are constants (and p will either be equal to n or m)[refer to the example above - second line in the factoring part]. Now, the factors will be of the form: (x + n)(x + m). Click here to see some examples.^{2}+ bx + c (a is the coefficient of x^{2}, b is the coefficient of x and c is a constant): 1) Multiply a and c (remember the sign) 2) Write down b (remember the sign) 3) Write down all the pairs of factors of the product of ac and see which pair add up to give you b (signs are important) 4) Rewrite the x term as the sum of the two terms with these numbers as coefficients.

*Perfect Squares*

Any expression of the form: x^{2} + 2ax + a^{2} is a *perfect square* because x^{2} + 2ax + a^{2} = (x + a)^{2} (i.e. it can be writen as (something)^{2}). [You can check to see that this is correct by expanding (x + a)^{2}]. To recognize an expression as a perfect square or not, you should first see if the constant term is a square number (i.e. can you take the square root of it and get an integer for an answer?). If so, see if the square root of it, multiplied by 2 (only if the coefficient of x^{2} is 1) gives you the coefficient of the linear term (the x term). If it does, the original expression may be factored into a *perfect square.* Click here to see an example. Note that if you do not recognize it as a perfect square, it can still be facoted by the decomposition method to get the same answer.

*Difference of Squares*

A *difference of squares* expression is one of the form: (something)^{2} - (something)^{2}. Mathematically, anything that looks like (a^{2}x^{2} - b^{2}) is a difference of squares and can be factored into (ax + b)(ax - b). Notice that the factors are identicle except for the sign. If you multiply out (ax + b)(ax - b), you get (ax)(ax) - abx + abx - b^{2}. The two middle terms ( -abx and + abx) cancel each other and you are left with (a^{2}x^{2} - b^{2}). Click here to see some examples.

**Note that not every expression is factorable using these above procedures. You know now that if a certain value, call it n, for x (or any variable that is in question) makes the whole expression zero, then

*Difference of Cubes*

A *difference of cubes* is an expression of the form: a^{3} - b^{3}. It can be factored into (a - b)(a^{2} + ab + b^{2}). To verify this, all you need to do is expand these factors:

(a - b)(a^{2} + ab + b^{2})

= a^{3} + a^{2}b + ab^{2} - a^{2}b - ab^{2} - b^{3}

= a^{3} - b^{2}

It is often difficult to memorize this factored expression so all you really need to remember is that if you ever see an expression like (a^{3} - b^{3}), then (a - b) is a factor. Once you remember that, use long division to find the remaining factor(s).

*Sum of Cubes*

A *sum of cubes* is an expression of the form: a^{3} + b^{3}. It can be factored into (a + b)(a^{2} - ab + b^{2}). To verify this, all you need to do is expand these factors:

(a + b)(a^{2} - ab + b^{2})

= a^{3} - a^{2}b + ab^{2} + a^{2}b - ab^{2} + b^{3}

= a^{3} + b^{2}

It is often difficult to memorize this factored expression so all you really need to remember is that if you ever see an expression like (a^{3} + b^{3}), then (a + b) is a factor. Once you remember that, use long division to find the remaining factor(s).

*Polynomials of Higher Degree*

When you first see a polynomial of degree greater than 2, that is not a difference or sum of cubes, there is no need to panic. Often, there will be a common factor in the expression which when factored out, will make the remaining expression a quadratic. For example: (9x^{4} - 2x^{3} + 10x^{2}), x^{2} is a common factor and thus 9x^{4} - 2x^{3} + 10x^{2} = x^{2}(9x^{2} - 2x + 10). The expression inside the brackets can now be factored using the decompostion method.
In some cases, you will be given one factor of a large expression and you will be need to find the remaining ones. A good way to do this is by using long division. Click here to see some examples.

If you are not given one of the factors, there is still hope in factoring it.
By now you should realize that if (x - k) is a factor [where k is a constant]
of some polynomial, then k must divide *exactly* into the *constant* term of the
polynomial. If the highest order term in a polynomial is not 1, then the factor is of the form (hx - k), where h must divide *exactly* into the coefficient of the highest order term, and k must divide *exactly* into the *constant* term of the polynomial. The possible values for k are the
factors of the constant term, and the possible values of h are the factors of
the leading coefficient (coefficient of the highest order term). You then must
use long division with these possible factors to see if they are indeed factors, and thus find the other factors. Click here to see some examples.

This method can be very time consuming and tedious if there is a lot of possible factors, so there is a shorter way to check if a possible factor is indeed a factor. Once you find the possible k and h values, and put them together in a possible factor form, you check which of them are factors by the following rules:

1) If (x - k) is a factor of the polynomial of interest, then substituting x = k into the polynomial gives an answer of zero. 2) If (x - k) is not a factor of the polynomial of interest, then substituting x = k into the polynomial does not give an answer of zero. 3) If (hx - k) is a factor of the polynomial of interest, then substituting x = k/h into the polynomial gives an answer of zero. 4) If (hx - k) is not a factor of the polynomial of interest, then substituting x = k/h into the polynomial does not give an answer of zero.

You can check all possible factors this way but it is quite possible that out of many of the possiblities, only one of them is correct, and the other factor that you need to find may be of a different form, such as (dx

^{2}+ ex - f). That means that you may waste time substituting without finding all the factors and that you will still need to perform long division. Once you have found one factor using these rules, it is a common practise to use long division to find all remaining factors. Click here to see an example.

*Summary of Methods for Factoring*

1) Take out any common factors.

2) Recognize if polynomial is (or isn't) a perfect square, a difference of squares, a difference of cubes or a sum of cubes.

3) If quadratic, try decomposition method.

4) If polynomial is higher than degree 2, find factors of the form (hx - k) or (x - k) by substituting x = k/h and or

x = k into the polynomial and then using long division.

4) Continue process until polynomial is fully factored.

5) Check the factors by multiplying them together - you should get the original polynomial if the factors are correct.