简单的数学题分解因式:1.x^4
分解因式1,原式=(X^4-10X^3+21X^2)+(14X^2-48X+18)
=X^2(X^2-10X+21)+(14X^2-48X+18)
=X^2(X-7)(X-3)+(14X-6)(X-3)
=(X-3)[X^2(X-7)+14X-6]
=(X-3)(X^3-7X^2+14X-6)
=(X-3)(X^3-3X^2-4X^2+14X-6)
=(X-3)[(X^3-3X^2)-(4X^2-14X+6)]
=(X-3)[X^2(X-3)-(4X-2)(X-3)]
=(X-3)(X-3)(X^2-4X+2)
分解因式2,
X^4-6X^2-7X-6
=X^4-4X^2-2X^2-7X-...全部
分解因式1,原式=(X^4-10X^3+21X^2)+(14X^2-48X+18)
=X^2(X^2-10X+21)+(14X^2-48X+18)
=X^2(X-7)(X-3)+(14X-6)(X-3)
=(X-3)[X^2(X-7)+14X-6]
=(X-3)(X^3-7X^2+14X-6)
=(X-3)(X^3-3X^2-4X^2+14X-6)
=(X-3)[(X^3-3X^2)-(4X^2-14X+6)]
=(X-3)[X^2(X-3)-(4X-2)(X-3)]
=(X-3)(X-3)(X^2-4X+2)
分解因式2,
X^4-6X^2-7X-6
=X^4-4X^2-2X^2-7X-6
=(X^4-4X^2)-(2X^2+7X+6)
=X^2(X^2-4)-(2X+3)(X+2)
=X^2(X+2)(X-2)-(2X+3)(X+2)
=(X+2)[X^2(X-2)-(2X+3)]
=(X+2)(X^3-2X^2-2X-3)
=(X+2)(X^3-3X^2+X^2-2X-3)
=(X+2)[(X^3-3X^2)+(X^2-2X-3)]
=(X+2)[X^2(X-3)+(X-3)(X+1)]
=(X+2)(X-3)(X^2+X+1)
3\3。
已知f(x)=x^5+3x+k^2能被x+1整除,求k之值
解:把原式分解因式,要使每项中都必须含有(x+1)
f(x)=x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+4x+4
=(x^5+x^4)-(x^4+x^3)+(x^3+x^2)-(x^2+x)+(4x+4)
=(x+1)x^4-(x+1)x^3+(x+1)x^2-(x+1)x+(x+1)4
=(x+1)(x^4-x^3+x^2-x+4)
因此k^2=4
得k=2
4。
已知x^2+x-1=0,求x^3+2x^2+2之值
解:由x^2+x-1=0
可得x^2+2x+1-x-2=0
(x+1)^2-(x+2)=0
(x+1)^2=x+2
x^3+2x^2+2
=x^3+2x^2+x-x+2
=x(x^2+2x+1)-(x-2)
=x(x+1)^2-(x-2) 这里用了上面的式子得出的结果进行等量代换
=x(x+2)-(x-2)
=x^2+x+2
=x^2+x-1+3
=0+3
=3
。
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