f3maths因式分解

1)y^3+y^2+y+1
= y^2(y+1)+(y+1)
= (y+1)(y^2+1)
2)4x^3+28x^2-5x-35
= 4x^2(x+7)-5(x+7)
= (x+7)(4x^2-5)
3)-c^2+2c+15
= -(c^2-2c-15)
= -(c-5)(c+3)
4)(6x-1)(x-3)+7
= 6x^2-18x-x+3+7
= 6x^2-19x+10
= (2x-5)(3x-2)
5)-12(r+4)^2+9(r+4)+3
= -3[4(r+4)^2-3(r+4)-1]
= -3(4r+16+1)(r+4-1)
= -3(4r+17)(r+3)
6)(x+4)^2-(x-4)^2
= (x+4+x-4)(x+4-x+4)
= (2x)(8)
= 16x
7)-3a^2+72+6a
= -3(a^2-2a-24)
= -3(a-6)(a+4)
8)27s^3+1
= (3s+1)(9s^2-3s+1)
9)81x^3-8
cannot be factorized

1)
y^3+y^2+y+1
=y^2(y+1)+(y+1)
=(y+1)(y^2+1)

2)
4x^3+28x^2-5x-35
=4x^2(x+7)-5(x+7)
=(x+7)(4x^2-5)

3)
-c^2+2c+15
=(c+3)(c-5)
4)
(6x-1)(x-3)+7
=6x^2-18x-x+3+7
=6x^2-19x+10
=(2x-5)(3x-2)

5)
-12(r+4)^2+9(r+4)+3
=[4(r+4)+1][(r+4)-1]
=(4r+16+1)(r+3)
=(4r+17)(r+3)

6)
(x+4)^2-(x-4)^2
=[(x+4)-(x-4)][(x+4)+(x-4)]
=16(2x)
=32

7)
-3a^2+72+6a
=-3a^2+6a+72
=(a+4)(a-6)

8)
27s^3+1
=(3s+1)(9s^2-3s+1)

9)
81x^3-8
can not

1)y^3+y^2+y+1
=y^2(y+1)+(y+1)
=(y^2+1)(y+1)
2)4x^3+28x^2-5x-35
=4x^2(x+7)-5(x+7)
=(4x^2-5)(x+7)
3)-c^2+2c+15
=-(c^2-2c-15)
=-(c-5)(c+3)
4)(6x-1)(x-3)+7
=6x^2-19x+10
=(2x-5)(3x-2)
5)-12(r+4)^2+9(r+4)+3
=-3(4(r+4)-3(r+4)-1)
=-3((r+4)-1)(4(r+4)+1)
=-3(r+3)(4r+17)
6)(x+4)^2-(x-4)^2
=[(x+4)+(x-4)][(x+4)-(x-4)]
=8x
7)-3a^2+72+6a
=-3(a^2-2a-24)
=-(a-6)(a+4)
8)27s^3+1
=(3s)^3+1^3
=(3s+1)[(3s)^2-(3s)(1)+1^2]
=(3s+1)(9s^2-3s+1)
9)81x^3-8
cannot be factorised since solving 81x^3-8=0, x=2/(3*cubicroot(3)), according to remainder theorem, this expression has no linear factor with integral coefficients.

y^3+y^2+y+1
=y^2(y+1)+(y+1)
=(y+1)(y^2+1)
4x^3+28x^2-5x-35
4x^2(x+7)-5(x+7)
=(x+7)(4x^2-5)

依2條f.2 數-.-
我f.2 咋....