Math 10034 ebook Sec 1 - Kent State University
Math 10034 ebook Sec 1.2 Factor by Grouping
Factoring a Four – Term Polynomial by Grouping
Step 1. Arrange the terms so that the first two terms have a common factor and the last two terms have a common factor.
Step 2. For each pair of terms, use the distributive property to factor out the pair’s greatest common factor.
Step 3. If there is now a common binomial factor, factor it out.
Step 4. If there are no common binomial factor in step 3, begin again, rearranging the terms differently. If no rearrangement leads to a common binomial factor, the polynomial cannot be factored.
Example 1) Factor 3x[pic]+ 4xy – 3x – 4y by grouping.
3x[pic]+ 4xy – 3x – 4y = x(3x + 4y) – (3x + 4y) = (3x + 4y)(x – 1)
Reminder: When factoring a polynomial, make sure the polynomial is written as a product. Do not write the final answer as a sum or difference of terms like: x(3x+4y) - (3x+4y) This form is not a factored form of the original polynomial. The factored form is the product: (3x + 4y)(x – 1)
Example 2) Factor ax – ab – 2bx + 2b[pic]by grouping.
ax – ab – 2bx + 2b[pic] = a(x – b) – 2b(x – b) = (x – b)(a -2b)
Sec 1.2 Exercises
Factor the four–term polynomial by grouping:
1. xy + 2x + 3y + 6 8. 3rs – s + 12r – 4
2. 2z – 8 + xz – 4x 9. 4b[pic] – 2bc – 7cd + 14bd
3. a[pic] + 4a[pic] + a + 4 10. x[pic] + 6x[pic] – 4x – 24
4. 4b[pic] – 8bc – 3b + 6c 11. a[pic] – 2a[pic] – 36a + 72
5. 3y – 5x + 15 – xy 12. x[pic] – 28 + 7x[pic] – 4x
6. 2x[pic] – x[pic]– 10x + 5 13. a[pic] – 45 – 9a + 5a[pic]
7. 12a[pic] – 42a[pic]– 4b + 14 14. -6 + 3x – 2x[pic] + x[pic]
Answers: Sec 1.2 Factor by Grouping
1. (y + 2)(x + 3)
2. (z – 4)(2 + x)[pic]
3. (a + 4)(a[pic]+ 1)
4. (b – 2c)(4b – 3)
5. (3 – x)(y + 5)
6. (2x –1)(x[pic]– 5)
7. 2(2b – 7)(3a[pic]– 1)
8. (3r – 1)(s + 4)
9. (2b – c)(2b + 7d)
10. (x + 6)(x + 2)(x – 2)
11. (a – 2) (a + 6)(a – 6)
12. (x + 7)(x + 2)(x – 2)
13. (a + 5)(a +3)(a – 3)
14. (x – 2)(x[pic]+ 3)
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