|
||||||||||||
|
||||||||||||
| ||||||||||||
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 x3 + 2x2 + 2x + 1 then grouping as (x3 + 1) + (2x2 + 2x) would yield
The quadratic x2 + 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)(x2 + x + 1) and not attempt to factor (x2 + x + 1). Harley
| ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |