Re: [中學] 國二因式分解
※ 引述《cecilia0305 (Cecilia)》之銘言：
: 1.求 (2y+1)(3y-1)(4y-1)(y+1)-36y^4的因式分解
= [(2y+1)(4y-1)][(y+1)(3y-1)]-36y^4
= (8y^2+2y-1)(3y^2+2y-1) - 36y^4
= 24y^4 + 11y^2(2y-1) + (2y-1)^2 - 36y^4
= -12y^4 + 11y^2(2y-1) + (2y-1)^2
12y^2 \/ (2y-1)
-1y^2 /\ (2y-1)
= (12y^2+2y-1)(-y^2+2y-1)
= -(12y^2+2y-1)(y-1)^2 (前一因式由一次因式檢驗法知無整係數一次因式)
: 2.求(x+y)(x+1)(y+1)+xy的因式分解
= (x+y)(xy+x+y+1)+xy
= (x+y)^2 + (x+y)(xy+1) + xy
(x+y) \/ xy
(x+y) /\ 1
= (x+y+xy)(x+y+1)
: 高雄某私立國中講義題目阿> <
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