因式分解^^

您好，我是 lop，高興能解答您的問題。

(1a)

9x^3 - 6x^2 + x
= x(9x^2-6x+1)
= x[(9x^2-3x) - (3x-1)]
= x[3x(3x-1) - 1(3x-1)]
= x(3x-1)(3x-1)
= x(3x-1)^2

(1b)

5x^2+20xy+20y^2
= 5(x^2+4xy+4y^2)
= 5[(x^2+2xy) + (2xy+4y^2)]
= 5[x(x+2y) + 2y(x+2y)]
= 5(x+2y)(x+2y)
= 5(x+2y)^2

(2a)

(y-3)^2+8(y-3)+16
= [(y-3)^2+4(y-3)] + [4(y-3)+16]
= (y-3)[(y-3)+4] + 4[(y-3)+4]
= (y-3+4)(y-3+4)
= (y+1)(y+1)
= (y+1)^2

(2b)

4(r+s)^2 + 20(r+s) + 25
= [4(r+s)^2+10(r+s)] + [10(r+s)+25]
= 2(r+s)[2(r+s)+5] + 5[2(r+s)+5]
= [2(r+s)+5][2(r+s)+5]
= [2(r+s)+5]^2
= (2r+2s+5)^2

(3a)

9x^2 + 12x + 4
= (9x^2+6x) + (6x+4)
= 3x(3x+2) + 2(3x+2)
= (3x+2)(3x+2)
= (3x+2)^2

(b)

9x^2 + 12x + 4 - 49y^2
= (3x+2)^2 - (7y)^2
= (3x+2+7y)(3x+2-7y)註(1)
= (3x+7y+2)(3x-7y+2)

註：

(1) 緊記公式 x^2 - y^2 = (x+y)(x-y)

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

1b.
5x^2+20xy+20y^2
=5(x^2+4xy+4y^2)
=5(x+2y)^2

2a.
(y-3)^2+8(y-3)+16
=[(y-3)+4)]^2
=(y-3+4)^2
=(y-1)^2

2b.
4(r+s)^2+20(r+s)+25
=[2(r+s)+5]^2
=(2r+2s+5)^2

3a.
9x^2+12x+4
=(3x+2)^2

3b.
9x^2+12x+4-49y^2
=(3x+2)^2-49y^2
=[(3x+2)-7y][(3x+2)+7y]
=(3x+2-7y)(3x+2+7y)
=(3x-7y+2)(3x+7y+2)

(1a)

9x^3 - 6x^2 + x
= x(9x^2-6x+1)
= x[(9x^2-3x) - (3x-1)]
= x[3x(3x-1) - 1(3x-1)]
= x(3x-1)(3x-1)
= x(3x-1)^2

(b)

5x^2+20xy+20y^2
= 5(x^2+4xy+4y^2)
= 5[(x^2+2xy) + (2xy+4y^2)]
= 5[x(x+2y) + 2y(x+2y)]
= 5(x+2y)(x+2y)
= 5(x+2y)^2

(2a)

(y-3)^2+8(y-3)+16
= [(y-3)^2+4(y-3)] + [4(y-3)+16]
= (y-3)[(y-3)+4] + 4[(y-3)+4]
= (y-3+4)(y-3+4)
= (y+1)(y+1)
= (y+1)^2

(b)

4(r+s)^2 + 20(r+s) + 25
= [4(r+s)^2+10(r+s)] + [10(r+s)+25]
= 2(r+s)[2(r+s)+5] + 5[2(r+s)+5]
= [2(r+s)+5][2(r+s)+5]
= [2(r+s)+5]^2
= (2r+2s+5)^2

(3a)

9x^2 + 12x + 4
= (9x^2+6x) + (6x+4)
= 3x(3x+2) + 2(3x+2)
= (3x+2)(3x+2)
= (3x+2)^2

(b)

9x^2 + 12x + 4 - 49y^2
= (3x+2)^2 - (7y)^2
= (3x+2+7y)(3x+2-7y)註(1)
= (3x+7y+2)(3x-7y+2)