University:
California State University, Northridge
Course:
MATH 360 | Abstract Algebra I
Academic year:

2024
Views:

310

Pages:

2
Author:

shortiexessau3y

Factoring Factoring by grouping The only method we know of factoring a polynomial with 4 terms is factoring by grouping. To do this we group the first two terms together and the second two terms together placing a + inbetween them. Negatives should stay with the number they are next to. We factor out the greatest common factor of the first two terms, and the greatest common factor of the second two terms. At this point what remains in the parenthesis should be the same in both pairs. If not you have made an error, or the polynomial can not be factored. Once we have the same thing in each parenthesis, we factor that out front. Example: xy + 3y -5x - 15 first we group the first two terms together and the second two terms together placing a + inbetween. Note the - stays with the 5. (xy + 3y) + (-5x - 15) the common factor for the first pair is y, and for the second pair is -5, we now factor each or these out y(xy/y + 3y/y) + -5(-5x/-5 - 15/-5) = y(x + 3) - 5(x + 3) Now we have the same thing in each parenthesis so we can factor it out front (x+3)[y(x+3)/(x+3) -5(x + 3)/(x+3)] = (x + 3)(y - 5) and so our answer is (x+3)(y-5) Factor each of the following. 1. x3 + x2 + x + 1 2. 2x3 + 6x2 + 4x + 12 3. 10x2 + 5x -6xy - 3y #1 solution 1. x3 + x2 + x + 1 first we group the first two terms together and the second two terms together placing a + inbetween. (x3 + x2) + (x + 1) the common factor for the first pair is x2, and for the second pair is 1, we now factor each or these out x2 (x + 1) + 1(x + 1) Now we have the same thing in each parenthesis so we can factor it out front (x+1)[x2(x+1)/(x+1) + 1(x + 1)/(x+1)] = (x + 1)(x2 + 1) and so our answer is (x + 1)(x2 + 1) #2 solution 2. 2x3 + 6x2 + 4x + 12 Here we have a common factor for all 4 terms, 2, so we must factor it out first. 2(x3 + 3x2 + 2x + 6) Next, we group the first two terms together and the second two terms together placing a + inbetween. 2[(x3 + 3x2 )+ (2x + 6)] the common factor for the first pair is x2, and for the second pair is 2, we now factor each or these out 2[x2 (x + 3) + 2(x + 3)] Now we have the same thing in each parenthesis so we can factor it out front 2(x+3)[x2 + 2] and so our answer is 2(x + 3)(x2 + 2) #3 solution 3. 10x2 + 5x -6xy - 3y first we group the first two terms together and the second two terms together placing a + inbetween. Note the - stays with the 6. (10x2 + 5x) + (-6xy - 3y) the common factor for the first pair is 5x, and for the second pair is -3y, we now factor each or these out 5x(2x + 1) + -3y (2x + 1) Now we have the same thing in each parenthesis so we can factor it out front