ex3−4x2−7x+10=1=e0
⇒x3−4x2−7x+10=0
By trial and error method, we can find that x=1 is a solution.
Applying synthetic division
x=11−4−71001−3−101−3−100
⇒x3−4x2−7x+10=(x−1)(x2−3x−10)=0
⇒(x−1)(x−5)(x+2)=0
⇒x=−2,1,5
⇒A={−2,1,5}
B={x∈Z:(1−x)(1+x)(x−4)(x+2)≥0}
⇒B={−2,−1,1,2,3,4}
A∪B={−2,−1,1,2,3,4,5}
A∩B={−2,1}
n[(A∪B)×(A∩B)]
=n(A∪B)×n(A∩B)=7×2=14