Factoring

Sometimes a polynomial is in expanded or distributed form and needs to be in factored form. One method of factoring is called grouping. If we go back to the distributive law:

a(b+c) = ab+ac

We can apply this to

ac+ad+bc+bd

Notice that the terms with a are grouped together and then notice that the terms with b are grouped together. Now the distributive law can be applied in reverse:

a(c+d)+b(c+d)

Then (c+d) can be factored out:

(a+b)(c+d)

One way to factor is to look for common factors in all the terms. For example:

ab+ac+ad+ae

All these terms have a as a factor. The a's can be factored out

a(b+c+d+e)

Sometimes not all the terms have a common factor. In this case, the common factor is only factored out where it occurs.

ab+ac+ad+ef

Since a does not occcur in the last term. It does not get factored out there. Only the first three terms are factored.

a(b+c+d)+ef

Another way to factor is by looking at the coefficients and the constant in the polynomial. A coefficient is a non variable number that is multiplied with a variable in a term. For example:

x²+2x+1

The 2 in the middle term would be the coefficient.

In factoring x²+2x+1 we would look at the 2 and try to find a pair of numbers that would add up to produce the sum of 2. Also, we would look at the constant 1 and try to determine a pair of factors that would multiply to produce 1. Since 1*1 produces the constant 1 and 1+1 = 2, the pair of numbers 1 and 1 will be included in the factored form of x²+2x+1.

Since x²+2x+1 is quadratic, the factored form will have the form of (a+b)(c+d). Since x is squared, it factors into x*x.

At this point, x²+2x+1 can be written as:

x²+(1+1)x+1*1

The x in the middle term can be distributed to result in the following

x²+x+x+1

Factoring out x in the first two terms yields

x(x+1)+x+1 or

x*(x+1)+1*(x+1)

Factoring out (x+1) results in

(x+1)(x+1)

This can be checked by expanding

Combining like terms results in x²+2x+1 which is the original expression.