ページ1:

3-1 利用提公因式做因式分解
2018.11.13
則B&C為A的因式
AA BCA =BC
A為B&C的倍式
1
因式分解:將x的二次式寫成兩個一次式的業積
ex已知多項式20+3x+1可因式分解成(2x+1)(x+m),則m=
SOLO
(2x+3x+1)+(2x + 1) = x+1
.. m = | #
sol
(2x+1)(x+m)=2x²+(2m+1)x +m
=2x+3x+1
m = 1 #
ex.提出公园式做因式分解
(1) 2x²-6x=2x(x-3).
(3) 2 (3x-1)+(3x-1)=(3x-1) (6x+)
(1) = 2xxx - 2xx 3
= 2x(x-3)#
(2) = 4x(2x-1)+4x+2(x+3)
= 4x(2x-1)+4x(2x+6)
= 4x(4x+5)#
(3)=2(3x-1) (3x-1) + (3x-1) x 1
=(3x-1) (6x-2+1)
=(3x-1) (6x-1) #
(2) 4x (2x-1) +8x(x+3)= 4x(4x+5) 。
0
ex變號提出公因式做因式分解,(x-5)(x+)-(5-x)(x-2)=(x-5)(2x+1) •
= (x-5)(x+1)+(x-5) (x-2)
= (x-5)[(x+1)+(x-2)]
=(x-5)(2x-1) #
ex. 利用分组做因式分解, x² - 5x-ax+5a=(x-5(0-0)
= (x²-5x)-(ax-50)
= x(x-5)-a(x-5)
=(x-5)(x-a) #

ページ2:

2018.11.15.
ex重新分組做因式分解,G-2)x(x²-2a)=(x+2)(a-x)。
= ax-2x-x² + 2a
a(x+2)-x(x+2)
= (x+2) (a-x) *
P.112~122