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二项式的分解因式 Factoring binomials
Mathematics 八年级 Grade eight 云南省施甸二中 沈正零 Zhengling Shen
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教学目标 Objectives 1、掌握二项式分解因式的概念 the concept of binomial Factoring.
2、了解分解因式的意义,以及它与整式乘法的关系 the meaning of factorization and its relationship with the integral expression . 3、经历从分解因数到分解因式的类比过程 the analogy from the decomposition to Factored. 4、感受分解因式在解决相关问题中的作用 the role of factorization in resolving the related issues. 5、学会观察、分析,提高分析问题和解决问题的能力 Learn to observe, analyze, improve the skills of analysis and problem-solving skills .
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教学内容 Teaching content 内容Content:二项式因式分解binomial factorization
教学重点Teaching points : 1、如何提公因式to find a common factor. 2、平方差公式Squares formula. 3、在教学中要使学生掌握逆向思维和类比的数学思想方法 Analogy and reverse thinking 教学难点Teaching Difficulties : 1、如何找公因式to find common factor. 2、如何把二项式变形为平方差公式的结构形式 How to transfer the deformation of binomials to the squares formula structure
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教学方法 Teaching Methods 以学生探究发现和自主学习为主,教师讲授为辅
let students learn by themselves with assistance from the teacher 教具准备: 需要多媒体设备辅助教学 computer-assisted 教学时数: 一课时 Teaching time:45minutes
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教学过程 Teaching Process 请同学们归纳一下:什么叫公因式?
(一)教学准备 preparation 1、复习公因式的概念review the concept of Common Factor 想一想:下列各题中两个因式的公因式是多少? Think: what is the common factor of them? (1)993 与-99 (2)3x2与-3x (3)-6a3与-2a (1) 993 and -99 (2) 3x2 and-3x (3)-6a3 and-2a 请同学们归纳一下:什么叫公因式? summarize : What is common factor? 2、复习乘法公式中的平方差公式 Review formula for the difference of squares (a+b)(a-b)= 请同学们归纳一下平方差公式: ask a student to summarize the fomula
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(二)讲授新课 1、对于“993 -99能被100整除吗?”小明是这样想的: Can " be divided by 100 ?" Since 993 -99=99×(992-1)=99×9800=99×98×100 所以小明认为它能被100整除。 the answer is yes 师问:那么993 -99还能被哪些正整数整除呢? Teacher asked:Can 993 -99 be divided by other positive integers? 学生答:还能被98和99整除。 Student A: 98 and 99 师: 回答正确,那么要解决这类问题的关键是什么呢? Teacher: correct, what is the key to solve such a problem ? 生: 是要把一个数式化成几个数的积的形式 Student :The key is to change a number into the form of multiplication of several numbers. 。 师:对,这就是今天我们要学的二项式的分解因式 Yes, that is decomposition of binomials.
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2、做一做:<先把题目展示给学生,然后请同学起来回答
filling the blanks 试计算下列各式: 3x(x-1)=________ (m+4)(m-4)= ________ a(a+1)(a-1)= ________ 根据上面的算式填空: 3x2-3x=( )( ) m2-16=( )( ) a3-a=( )( )( )
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3、议一议 further discussion:<根据练习情况进一步讨论>
由a(a+1)(a-1)得到a3-a的变形是什么运算?由a3-a得到a(a+1)(a-1)的变形与这种运算有什么不同?你还能再举一些类似的例子加以说明吗? How do you get a (a +1) (a-1) from a3-a ? What is the difference between it and the opposite calculation, from a3-a to a (a +1) (a-1)? 师生共同总结:(经过议论之后) 由a(a+1)(a-1)得到a3-a是整式的乘法;由a3-a得到a(a+1)(a-1)的变形是整式乘法的逆运算。 We get a(a+1)(a-1) from a3-a through multiplication, but we get a (a +1) (a-1) from a3-a is just the opposite 师:把一个二项式化成几个整式的积的形式,这种变形叫做把这个二项式分解因式。 Teacher: the way we change binomials into a multiplication of several integral expression is factoring binomials.
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4、想一想: 师问:分解因式的一般步骤是什么?(教师引导学生共同总结) What are the general steps? 生答:(1)如果二项式的各项有公因式,那么先提公因式 If there are common binomials, they should be taken out first e.g.:993 -99, x-xy 等,etc (2)如果各项没有公因式,那么可以尝试运用平方差公式进行分解 If the there is no common factor, you can try to use the formula for the difference of squares e.g.:a2-b2, 25-36x2,x2-4y2 ,等 (3)分解因式,必须进行到每一个多项式都不能再分解为止 Do not stop until you pick out the last common factor e.g.:x3-x=x(x2-1)= x(x +1)(x-1)
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(三)、课堂练习 Practice 用线连一连:matching x2-y (3-5x)(3+5x) 9-25x y(x-y) xy-y (2a-1)(2a+1) 4a2- (x+y)(x-y)
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课堂小结 Summary 1.这节课你有哪些收获? What have you learned from this lesson ?
2.你学到了哪些数学方法和数学思想? What mathematical methods have you learned from this lesson ?
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结果与分析 这节课充分调动了学生的积极性,主动性,以学生学习、探究为主,教师讲授为辅,在一问一答,互动讨论的情境下循序渐进地开展教学,既培养了学生的自主探究能力,也培养了学生学习数学的兴趣,体现了再创造,再发现的过程;在活动过程中学到了知识,提高了能力。 It makes the students active. With the assistance from their teachers, those students learn in interactive discussion or answering questions. It helps to cultivate their ability of independent learning.
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