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把下列各式分解因式:(1)x4-7x2-18;(2)m4n+m3n2-m2n3-mn4.

题目详情
把下列各式分解因式:
(1)x4-7x2-18;    (2)m4n+m3n2-m2n3-mn4

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▼优质解答
答案和解析
(1)由题意知
x4−7x2−18
=(x2−9)(x2+2)
=(x+3)(x−3)(x2+2)

=(x+3)(x-3)(x+
2
i)(x-
2
i)
(2)由题意知
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
x4−7x2−18
=(x2−9)(x2+2)
=(x+3)(x−3)(x2+2)
x4−7x2−18
=(x2−9)(x2+2)
=(x+3)(x−3)(x2+2)
x4−7x2−18
=(x2−9)(x2+2)
=(x+3)(x−3)(x2+2)
x4−7x2−18
=(x2−9)(x2+2)
=(x+3)(x−3)(x2+2)
x4−7x2−18x4−7x2−18x4−7x2−184−7x2−182−18=(x2−9)(x2+2)=(x2−9)(x2+2)=(x2−9)(x2+2)2−9)(x2+2)2+2)=(x+3)(x−3)(x2+2)=(x+3)(x−3)(x2+2)=(x+3)(x−3)(x2+2)2+2)
=(x+3)(x-3)(x+
2
i)(x-
2
i)
(2)由题意知
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
2
2
22i)(x-
2
i)
(2)由题意知
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
2
2
22i)
(2)由题意知
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
m4n+m3n2−m2n3−mn4
=mn(m3+m2n−mn2−n3)
=mn[(m3+m2n)−(mn2+n3)]
=mn[m2(m+n)−n2(m+n)]
=mn(m+n)(m2−n2)
=mn(m+n)2(m−n)
m4n+m3n2−m2n3−mn4m4n+m3n2−m2n3−mn4m4n+m3n2−m2n3−mn44n+m3n2−m2n3−mn43n2−m2n3−mn42−m2n3−mn42n3−mn43−mn44=mn(m3+m2n−mn2−n3)=mn(m3+m2n−mn2−n3)=mn(m3+m2n−mn2−n3)3+m2n−mn2−n3)2n−mn2−n3)2−n3)3)=mn[(m3+m2n)−(mn2+n3)]=mn[(m3+m2n)−(mn2+n3)]=mn[(m3+m2n)−(mn2+n3)]3+m2n)−(mn2+n3)]2n)−(mn2+n3)]2+n3)]3)]=mn[m2(m+n)−n2(m+n)]=mn[m2(m+n)−n2(m+n)]=mn[m2(m+n)−n2(m+n)]2(m+n)−n2(m+n)]2(m+n)]=mn(m+n)(m2−n2)=mn(m+n)(m2−n2)=mn(m+n)(m2−n2)2−n2)2)=mn(m+n)2(m−n)=mn(m+n)2(m−n)=mn(m+n)2(m−n)2(m−n)