Saturday, August 31, 2013

find the zeros fro g(x), g(x) = 4(x + 2)^5 (x-3)^3 + 5(x-3)^4 (x+2)^4

by grouping 
4%28x+%2B+2%29%5E5+%28x-3%29%5E3+%2B+5%28x-3%29%5E4+%28x%2B2%29%5E4 
becomes 
%28x%2B2%29%5E4%28x-3%29%5E3%284%28x%2B2%29%2B5%28x-3%29%29 

which simplifies to 
%28x%2B2%29%5E4%28x-3%29%5E3%289x-7%29 
so there are 3 unique solutions: 
x=-2 (multiplicity 4)
x=3 (multiplicity 3)
x=7/9

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