



 
David, Factoring this quintic polynomial involves seeing a pattern. If you group the terms
You can see that each grouping has a factor of x + 2 and extracting this common factor yields
for some numbers a, b and c. This can then be factored again to finally give h(x) as a product of linear factors and hence the zeros are all real. If the expression had been x^{3} + 2x^{2} + 2x + 1 then grouping as (x^{3} + 1) + (2x^{2} + 2x) would yield
The quadratic x^{2} + x + 1 has imaginary roots but the instruction "Leave factors with imaginary zeros in quadratic form." means that you should leave the factorization as (x + 1)(x^{2} + x + 1) and not attempt to factor (x^{2} + x + 1). Harley
 


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