Maths 問題~F.2 Factorization

1. (x－3)^2－4x^2

=(x－3－2x)(x－3+2x)

=(－x－3)(3x－3)

=－3(x+3)(x－1)



2. (x^2－y^2) +2x－2y

=(x－y)(x+y)+2(x－y)

=(x－y)(x+y+2)



3. 4(x+y)^2 +12(x+y)+9

=(2x+2y)^2+2*(2x+2y)*3+3^2

=(2x+2y+3)^2



4. (x^2－6x + 9)－4y^2

=(x－3)^2－(2y)^2

=(x－2y－3)(x+2y－3)



5. Using the result of 9x^2 + 12xy + 4y^2，factorize 9x^2 +12xy－21y^2

Sol

9x^2+12xy+4y^2=(3x+2y)^2

9x^2+12xy－21y^2

=(9x^2+12xy+4y^2)－25y^2

=(3x+2y)^2－(5y)^2

=(3x+2y+5y)(3x+2y－5y)

=(3x+7y)(3x－3y)

=3(3x+7y)(x－y)

1. (x-3)^2 - 4x^2
=(x-3+2x)(x-3-2x)
=(3x-3)(-x-3)
=-3(x-3)(x+1)
解決No.1要用identity(恆等式):
(a^2)-(b^2)=(a+b)(a-b)
然後再抽3 and -

2.(x^2-y^2) +2x -2y
=(x+y)(x-y)+2(x-y)
=(x+y+2)(x-y)

3.4(x+y)^2 +12(x+y)+9
=[2(x+y)]^2+2[2(x+y)](3)+(3)^2
=[2(x+y)+3]^2
=(2x+2y+3)^2

諗唔明就let a=(x+y) 代=(x+y)
4(x+y)^2 +12(x+y)+9
=4a^2+12a+9
=(2a)^2+2(2a)(3)+(3)^2
=(2a+3)^2
=[2(x+y)+3]^2
=(2x+2y+3)^2

4. (x^2 - 6x + 9)- 4y^2
=[(x)^2-2(x)(3)+(3)^2]-(2y)^2
=(x-3)^2-(2y)^2
=(x-2y-3)(x+2y-3)

5.By Using the result of 9x^2 + 12xy + 4y^2 ,
factorize 9x^2 +12xy- 21y^2

9x^2 + 12xy + 4y^2
=(3x)^2+2(3x)(2y)+(2y)^2
=(3x+2y)^2

9x^2 +12xy- 21y^2
=9x^2 + 12xy + 4y^2-25y^2
=(3x+2y)^2-(5y)^2
=(3x+2y-5y)(3x+2y+5y)
=(3x-3y)(3x+7y)
=3(x-y)(3x+7y)

1. (x-3)^2 - 4x^2
=x^2-2(3)(x)+(-3)^2-(2x)^2
=x^2-6x+9-(2x)^2
=-3x^2-6x+9
=-3(x^2+2x-3)
=-3(x+3)(x-1)

2. (x^2-y^2) +2x -2y
=(x+y)(x-y)+2(x-y)
=(x-y)(x+y+2)

4. (x^2 - 6x + 9)- 4y^2
=(x-3)^2-(2y)^2
=(x-3+2y)(x-3-2y)