數學作業 因式分解 超急

題目太多, 作答需時, 請等等。

2014-08-08 12:07:36 補充：
1. 因 2x^3+3x^2-5x-6 最高次方是3, 而 x+1 最高次方是1,
所以, 2x^3+3x^2-5x-6 另一個因式最高次方是 3-1=2
2x^3+3x^2-5x-6
= (x+1)(ax^2+bx+c)
= ax^3+bx^2+cx+ax^2+bx+c
= ax^3+(b+a)x^2+(c+b)x+c
所以, a=2, b=1, c= -6
2x^3+3x^2-5x-6
= (x+1)(2x^2+x-6)
= (x+1)(2x-3)(x+2) (ps.(2x^2+x-6) = (2x-3)(x+2)的方法是 十字交乘法)

2. (2x-1)^2-(x+1)^2 = 45
[(2x-1)+(x+1)] [(2x-1)-(x+1)] = 45
(3x)(x-2) = 45
3x^2-6x-45 = 0
x^2-2x-15 = 0
(x-5)(x+3) = 0
x= 5 或 -3
(ps. 如果是 (2x-1)^2+(x+1)^2 = 45的話, 就要展開再重組)

3.(1). 13x^2(x+5y)-7xy(x+5y)-6y^2(x+5y)
= (x+5y)(13x^2-7xy-6y^2)
= (x+5y)(13x+6y)(x-y)
(ps. 小提示(13x^2-7xy-6y^2) = (13x+ay)(x+by), 13是質數, 只有1和13是因數)

3(2). x^2y-x^2+3xy-2x+2y-1
= (x^2y+3xy+2y)+(-x^2-2x-1) (ps. 有y的組在一起, 剩下的一組)
= y(x^2+3x+2)-(x^2+2x+1)
= y(x+1)(x+2)-(x+1)^2
= (x+1)[(x+2)y-(x+1)] (或 (x+1)[(y-1)x+2y-1] )

4.(1). (x+1)^2-(x+1)-12
= [(x+1)-4][(x+1)+3]
(ps. 把(x+1)當作一個整體, 或(x+1)=y, y^2-y-12會容易想一些)
= (x-3)(x+4)

4.(2). (x^2+4x-5)(x^22+4x+3)-105 應該題目錯了,
應該是 (x^2+4x-5)(x^2+4x+3)-105
= x^4+4x^3+3x^2+4x^3+16x^2+12x-5x^2-20x-15-105
= x^4+8x^3+14x^2-8x-120
= (x-2)(ax^3+bx^2+cx+d)
(ps. (2)^4+8(2)^3+14(2)^2-8(2)-120 = 0, (x-2)是其中一個因式)
= ax^4+bx^3+cx^2+dx-2ax^3-2bx^2-2cx-2d
= ax^4+(b-2a)x^3+(c-2b)x^2+(d-2c)-2d
所以, a=1, b=10, c=34, d=60
(x^2+4x-5)(x^2+4x+3)-105
= (x-2)(x^3+10x^2+34x+60)
= (x-2)(x+6)(x^2+4x+10)
(ps. (-6)^4+8(-6)^3+14(-6)^2-8(-6)-120 = 0, (x-2)是其中一個因式)

5.(1). (x+a)(x-2) = x^2+bx+2
x^2+(a-2)x-2a = x^2+bx+2
所以, (a-2) = b, -2a = 2
a= -1, b= -3

5.(2). A = 0,
(x+a)(x-2) = 0
(x-1)(x-2) = 0
x = 1 或 2

6.(1). x^2-6x+8=0
(x-2)(x-4) = 0
x = 2 或 4
另一個根是4

6.(2). a(x-2)(x+1)+a(x-2) = 0
a(x-2)(x+1+1) = 0
a(x-2)(x+2) = 0
x = -2 或 2

7. a(a-5) = 14
a^2 - 5a - 14 = 0
(a-7)(a+2) = 0
a = -2 或 7
如果a = -2
(a+2)/(a-4) = (-2+2)/(-2-4) = 0
如果a = 7, (a+2)/(a-4) = (7+2)/(7-4) = 9/3 = 3

2014-08-08 12:08:21 補充：
剩下的在意見區

2014-08-08 12:14:28 補充：
8. 108x^2+ax-77 = (bx-7)(12x+c)
108x^2+ax-77 = 12bx^2+(bc-84)x-7c

12b = 108
b = 9

-7c = -77
c = 11

a = bc-84 = (9)(11)-84 = 15

a-b-c = 15-9-11 = -5

2014-08-08 12:17:19 補充：
9. 12x^2-9x/5 = 3/10
40x^2 - 6x - 1 = 0
(4x-1)(10x+1) = 0
x= 1/4 或 -1/10

2014-08-08 12:19:07 補充：
10.(1). (x-4)(x+5) = -8
x^2+x-20+8 = 0
x^2+x-12 = 0
(x+4)(x-3) = 0
x = -4 或 3

2014-08-08 12:22:22 補充：
10.(2) 如果x = -4,
(-3x+5)/(-x+1) = [-3(-4)+5]/[-(-4)+1] = 17/5
如果x = 3,
(-3x+5)/(-x+1) = [-3(3)+5]/[-(3)+1] = -4/(-2) = 2

2014-08-08 12:25:52 補充：
11. ax^2+bx-5 = 0
(x-1/2)(x-(-5/2)) = 0
(2x-1)(2x+5) = 0
4x^2+8x-5 = 0

所以, a=4, b=8

2014-08-08 12:46:51 補充：
12. x^2+kx-15 = (ax+b)(cx+d), a,b,c,d是整數
如果 x^2+kx-15 = 0
(ax+b)(cx+d) = 0
x= -b/a 或 -d/c
x = [-k+√(k^2-4(1)(-15))] / [2(1)] 或 [-k-√(k^2-4(1)(-15))] / [2(1)]
x = [-k+√(k^2+60)] / 2 或 [-k-√(k^2+60)] / 2

2014-08-08 12:55:02 補充：
即 -b/a = [-k+√(k^2+60)] / 2 (或 [-k-√(k^2+60)] / 2 )
-d/c = [-k-√(k^2+60)] / 2 (或 [-k+√(k^2+60)] / 2 )
因b,d是整數, -k+√(k^2+60) 和 -k-√(k^2+60) 是整數
即√(k^2+60) 是整數 (或k^2+60是平方數)
設k = 2(題目10個數字的第1個)
√(k^2+60) = √(2^2+60) = √64 = 8(是整數)
所以k = 2符合題目要求

2014-08-08 12:59:42 補充：
再繼續試剩下的數字,
(ps. 如果那個正數是符合題目要求, 相對應的負數也一定符合題目要求,
所以不用試題目中的負數,
因為無論k是正數還是負數, k^2也是一樣, 即√(k^2+60)也是一樣)
k = 2, -2, 14, -14符合題目要求,
有4個是可以是k的值