walisunita
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- Aug 28, 2017
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Please help in calcuiating the zeroes of the cubic polynomial ... ( X3 - 3 )
Here's one approach.Please help in calcuiating the zeroes of the cubic polynomial ... ( X3 - 3 )
Here's one approach.
Let C = cube root of 3
Use C to rewrite your polynomial as a Difference of Cubes, then factor it.
Using the Zero-Product Property, you'll get one solution from the linear factor, and you'll get two solutions from the quadratic factor, each in terms of C.
Finish, by replacing symbol C with the cube root of 3.
If you need more help, please show us what you tried or thought about. :cool:
I don't understand. If we are dealing with real zeroes, either approach will easily find the only real zero because the determinant of the quadratic will be negative. If we are dealing with complex zeroes, only one approach will easily find all three zeroes.If X is complex this approach is correct. If X is real, there will be no solutions from the quadratic factor.
If X is real here's another approach.
\(\displaystyle X^3 -3=0\)
\(\displaystyle X^3 =3 \) ... adding 3 to both sides
\(\displaystyle X = \sqrt[3]{3}\) ... taking cube root of both sides
What method have you been taught in the class?Please help in calculating the zeroes of the cubic polynomial ... ( X3 - 3 )
I'm sorry but what don't you understand? Yes both methods will find the only real solution. I was just pointing out that one is method is simpler than the other if you are looking for only the real solutions.I don't understand. If we are dealing with real zeroes, either approach will easily find the only real zero because the determinant of the quadratic will be negative. If we are dealing with complex zeroes, only one approach will easily find all three zeroes.