finding the zeroes of the cubic polynomial x^3 - 3

[walisunita]

Please help in calcuiating the zeroes of the cubic polynomial ... ( X³ - 3 )

 

[mmm4444bot]

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:

 

[Harry_the_cat]

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:

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

 

[JeffM]

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

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.

 

[Deleted member 4993]

Please help in calculating the zeroes of the cubic polynomial ... ( X³ - 3 )

What method have you been taught in the class?

Are you supposed estimate the value with certain decimal number?

 

[Harry_the_cat]

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.

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.