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huyunxiang 幼苗
共回答了303个问题 举报
设x²+8x+3=m
∵(3x²+24x+7)(2x²+16x+15)+14
=6m²+23m-4
=﹙6m+1﹚﹙m-4﹚
①6m+1=6﹙x²+8x+3﹚+1=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚
②m-4=x²+8x-1=﹙x+4-√17﹚﹙x+4+√17﹚
∴ (3...
∵(3x²+24x+7)(2x²+16x+15)+14
=6m²+23m-4
=﹙6m+1﹚﹙m-4﹚
①6m+1=6﹙x²+8x+3﹚+1=﹙6x﹢24-√462﹚﹙x﹢﹤24+√462﹥/6﹚
②m-4=x²+8x-1=﹙x+4-√17﹚﹙x+4+√17﹚
∴ (3...
1年前
1
共回答了1627个问题 举报
f(x)= (3x²+24x+7)(2x²+16x+15)+14
f(-1) = 0
f(-7)= 0
f(x) = (x+1)(x+7)(6x^2+ax+b)
coef .of const.
7b= 119, =>b=17
put x= 1
1136= 16(6+a+17)
a= 48
f(x) = (x+1)(x+7)(6x^2+48x+17)
= -1952/5
f(x) = (x-1)(x-7)(6x^2-(1952/5)x+17)
f(-1) = 0
f(-7)= 0
f(x) = (x+1)(x+7)(6x^2+ax+b)
coef .of const.
7b= 119, =>b=17
put x= 1
1136= 16(6+a+17)
a= 48
f(x) = (x+1)(x+7)(6x^2+48x+17)
= -1952/5
f(x) = (x-1)(x-7)(6x^2-(1952/5)x+17)
1年前
0
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