How do you factor polynomials?

Answer 1

A few examples..

** First, taking out a common factor:

1:

x3 + 3x

x(x2 + 3) -This is factored because you have taken an "x" away from both terms... to put it back into its original form just multiply the "x" by x2 and 3.

2:

6x2 + 16x + 8

2(3x2 + 8x + 4) -All of the numbers were divisible by 2 so you can take 2 out of all of the terms.

** Factoring a Difference of Squares:

x2 - 16 = (x)2 - 42 = (x - 4)(x + 4).

** An 'easy' trinomial (the coefficient of x is 1):

x2 + 5x + 6 = (x + 3)(x + 2). -Notice that 2+3=5 and (2)(3) =6.

** A 'hard' trinomial (the coefficient of x2 is not 1):

8x2 - 2x - 21 = (2x + 3)(4x - 7). -There are several ways of organizing a trial and error process to factor a case like this.

** Factoring by grouping (Check for this when there are 4 terms.)

6x3 - 2x2 + 15x - 5 = 2x2(3x - 1) + 5(3x - 1) = (2x2 + 5)(3x - 1).

Answer 2

A few examples..

** First, taking out a common factor:

1:

x3 + 3x

x(x2 + 3) -This is factored because you have taken an "x" away from both terms... to put it back into its original form just multiply the "x" by x2 and 3.

2:

6x2 + 16x + 8

2(3x2 + 8x + 4) -All of the numbers were divisible by 2 so you can take 2 out of all of the terms.

** Factoring a Difference of Squares:

x2 - 16 = (x)2 - 42 = (x - 4)(x + 4).

** An 'easy' trinomial (the coefficient of x is 1):

x2 + 5x + 6 = (x + 3)(x + 2). -Notice that 2+3=5 and (2)(3) =6.

** A 'hard' trinomial (the coefficient of x2 is not 1):

8x2 - 2x - 21 = (2x + 3)(4x - 7). -There are several ways of organizing a trial and error process to factor a case like this.

** Factoring by grouping (Check for this when there are 4 terms.)

6x3 - 2x2 + 15x - 5 = 2x2(3x - 1) + 5(3x - 1) = (2x2 + 5)(3x - 1).

These are the basic types of factoring in Algebra I. In more advanced courses, there are methods for factoring more complicated polynomials.