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Two of the zeros are `sqrt 3` and `-sqrt3` which means that two factors of the polynomial are `x-sqrt3` and `x+sqrt3` . The remaining two factors must then be `x-r_1` and `x-r_2` where `r_1` and `r_2` are the other roots. By expanding these factors, we get:
But we are told that the sum of the other two zeros are -1 and their product is -2, which means that this quadratic becomes:
Multiplying by the first two factors, we get the polynomial up to an arbitrary scaling factor to get
where `a ne 0` is any real number.
The polynomial is `f(x)=a(x^4+x^3-5x^2-3x+6)` .
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