Question

几道初二分解因式题···

x²+14x+40 2x²-7x-15 3x+x-24 10y²-13y-3
知道的写下步骤啊！

Answer 1

解：x²+14x+40
=x²+14x+40+9-9
=(x+7)²-9
=([(x+7)+3][(x+7)-3]

2x²-7x-15
=2(x²-7x/2-15/2 )
=2(x²-7x/2+49/16-49/16-15/2)
=2[(x-7/4)²-169/16]
=2[x-7/4)+13/4][(x-7/4)-13/4]

3x²+x-24
=3(x²+x/3-8)
=3(x²+x/3+1/36-1/36-8)
=3[(x+1/6)²-289/36]
=3[(x+1/6)²+17/6][(x+1/6)²-17/6]

10y²-13y-3
=10(y²-13y/10-3/10)
=10(y²-13y/10+169/400-169/400-3/10)
=10[(y-13/20)²-289/400]
=10[(y-13/20)+17/20][(y-13/20)-17/20]

请采纳！0(∩_∩)0

Answer 2

x²+14x+49-9
=(x+7)²-9
=[(x+7)+3][(x+7)-3]
=(x+10)(x+4)

2x²-7x-15
=(2x+3)(x-5)（十字相乘法）

3x+x-24（如果题目没错）=4x-24=4(x-6)
如果是3x²+x-24=(3x-8)(x+3)

10y²-13y-3=(2y-3)(5y+1)

Answer 3

x²+14x+40=x^2+(10+4)x+10*4=(x+4)(x+10)
2x²-7x-15=2x^2-(10-3)x-3*5=(2x+3)(x-5)
3x^2+x-24=3x^2+(9-8)x-3*8=(3x-8)(x+3)
10y²-13y-3=2*5x^2-(15-2)x-3=(5x+1)(2x-3)