University:
California State University, Northridge
Course:
MATH 102 | Pre-Calculus I
Academic year:

2024
Views:

454

Pages:

2
Author:

obratnoj4w0

Factoring Factoring completely When factoring a polynomial, in order for the answer to be correct we wish for the polynomial to be factored completely. Sometimes we must factor something more than once in order for it to be completely factored. A common mistake is not to factor out a common factor, this is why we do this first. Example: x6 - 64 This problem is both a difference of squares, and a difference of cubes In this case we wish to first factor it as a difference of squares the square root of x6 is x3, and the square root of 64 is 8 so we get (x3 - 8) (x3 + 8) Note that our first factor is a difference of cubes, and our second factor is a sum of cubes, we must factor each of these. (x3 - 8) factors into (x-2)(x2 + 2x + 4) (x3 + 8) factors into (x + 2)(x2 - 2x + 4) so our answer is (x-2)(x2 + 2x + 4)(x + 2)(x2 - 2x + 4) Factor each of the following. 1. x3 + 3x2 - 4x - 12 2. x12 - y12 3. 10x2 + 5x -15 #1 solution 1. x3 + 3x2 - 4x - 12 Here we have no common factor We have four terms which tells us to use factoring by grouping (x3 + 3x2 ) + (-4x - 12) x2 (x + 3) + -4(x + 3) (x + 3)(x2 - 4) x2 - 4 is a difference of squares, it can be factored x2 - 4 = (x-2)(x + 2) so our answer is (x + 3)(x - 2)(x + 2) #2 solution 2. x12 - y12 This problem is both a difference of squares, and a difference of cubes In this case we wish to first factor it as a difference of squares the square root of x12 is x6, and the square root of y12 is y6 so we get (x6- y6) (x6 + y6) Note that our first factor is a difference of cubes and a difference of squares, and our second factor is a sum of cubes, we must factor each of these. (x6- y6) we factor first as a difference of squares the square root of x6 is x3, and the square root of y6 is y3 so we get (x3- y3) (x3 + y3) Note that our first factor is a difference of cubes and a difference of squares, and our second factor is a sum of cubes, we must factor each of these. (x3- y3)factors into (x - y)(x2 + xy + y2) (x3 + y3)factors into (x + y)(x2 - xy + y2) Now we still need to factor (x6 + y6) (x6 + y6) factors into (x2 + y2)(x4 - x2y2 + y4) so our answer is (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)(x2 + y2)(x4 - x2y2 + y4) #3 solution 3. 10x2 + 5x -15 Here we have a common factor of 5 which must be factored out first 5(2x2 + x - 3) Since we have a trinomial we can use the ac method ac = -6 and b = 1 since b is positive and ac is negative, the larger factor is positive and the smaller is negative the factors of -6 are -1,6 -2,3 -2 and 3 add to be 1 so we use -2 and 3to break up the middle term 2 2x - 2x + 3x - 3; note: we have not changed the problem since -2x + 3x =1x Now we can use factoring by grouping (2x2 -2x) + (3x - 3) 2x(x - 1) + 3(x - 1) and our answer is 5(x -1)(2x + 3)