การแยกตัวประกอบของพหุนาม(โดยวิธีการจัดกลุ่มแล้วดึงตัวร่วม)

If a polynomial has four terms, you may be able to factor by grouping. factor out the GCF of the first two terms and the GCF of the second two terms. If the expressions in parentheses match, you can factor by grouping:

ac + ad + bc + bd
a(c + d) + b(c + d)
(a + b)(c + d)

Example 1          

8z³ + 7z² – 16z – 14
z²(8z + 7) – 2(8z + 7)  Factor by grouping
(z² – 2)(8z + 7)  Apply the distributive property

ตัวอย่างที่ 1 จงแยกตัวประกอบ ของ m(n+3)+5(n+3)

วิธีทำ m(n+3)+5(n+3)     =(m+5)(n+3)

ดังนั้น m(n+3)+5(n+3)     =(m+5)(n+3)
........................................................................................................................................

ตัวอย่างที่ 2 จงแยกตัวประกอบ ของ ax+by+bx+ay

วิธีทำ ax+by+bx+ay                         =(ax+ay)+(bx+by)

                                                        =(x+y)a+(x+y)b

                                                        =(a+b) (x+y)

ดังนั้น ax+by+bx+ay =(a+b)(x+y)
........................................................................................................................................

ตัวอย่างที่ 3 จงแยกตัวประกอบ ของ ab²-cb²-6a+6c

วิธีทำ ab²-cb²-6a+6c                        =(ab²-cb²)+(-6a+6c)

                                                        =(a-c)b²+(-a+c)6

                                                        =(a-c)b²+(a-c)(-6)

                                                        =(a-c)(b²-6)

ดังนั้น ax+by+bx+ay =(a-c)(b²-6)
..........................................................................................................................................

ตัวอย่างที่ 4 จงแยกตัวประกอบ ของ x³-x³z+y²z-y²

วิธีทำ x³-x³z+y²z-y²                          =(x³-x³z)+(y²z-y²)

                                                                =(1-z)x³+(z-1)y²    =(-z+1)x³+(z-1)y²            

                                                                =(z-1)(-x³)+(z-1)y²

                                                                =(z-1)(-x³+y²)

ดังนั้น ax+by+bx+ay =(z-1)(-x³+y²)
..........................................................................................................................................

เฉลยแบบฝึกหัด 1.1 ข

จงแยกตัวประกอบของพหุนามต่อไปนี้(โดยการจัดกลุ่มดึงตัวร่วม)

1.m(n+3)+5(n+3)            
=(n+3)(m+5)
........................................................................................................................................

2.(x+y)z-(x+y)                 
= (x+y)(z-1)
........................................................................................................................................

3.4t(a+b)-s(a+b)                             
=(a+b)(4t-s)
........................................................................................................................................

4.(4y²+3)y+6(4y²+3)      
=(4y²+3)(y+6)
........................................................................................................................................

5.a(b-3c)+x(b-3c)           
=(b-3c)(a+x)
.......................................................................................................................................

6.ax+by+bx+ay               
=(ax+bx)+(ay+by)
=(a+b)(x+y)
.......................................................................................................................................

7.5a-10x+ab-2bx            
=(5a-10x)+(ab-2bx)
=(a-2x)(5+b)
.......................................................................................................................................

8.na+3b+nb+3a                              
=(na+nb)+(3a+3b)
=(a+b)(n+3)
.......................................................................................................................................

9.xy-st-xt-sy                     
=(xy+sy)-(xt+st)
=(x+s)(y-t)
.......................................................................................................................................

10.n²m+n²p-8m-8p       
=(n²m-8m)+(n²p-8p)
=(m+p)(n²-8)
.......................................................................................................................................

11.ab²-cb²-6a+6c           
=(ab²-cb²)-(6a-6c)
=(a-c)(b²-6)
.......................................................................................................................................

12.2x³-x+14x²-7                          
=(2x³-x)+(14x²-7)
=(2x²-1)(x+7)
.......................................................................................................................................

13.a²-2b-5a³+10ab         
=(a²-2b)-(5a³-10ab)
=(a²-2b)(1-5a)
.......................................................................................................................................

14.x³-x³z+y²z-y²                           
=(x³-x³z)-(y²-y²z)
=(1-z)(x³-y²)
........................................................................................................................................