Question

*20 points* Find all the zeroes of the polynomial function f(x)=x^3-5x^2+6x-30. If you use synthetic division, show all three lines of numbers.

Answer 1

`x_(1) =5\\x_(2)=i√(6)\\x_(3)=-i√(6)\\`

Explanation:

Unlike quadratic equations, cubic equations might be harder to solve. Therefore, one useful approach is to group and factor terms:

`x^3-5x^2+6x-30=0\\x^2(x-5)+6(x-5)=0`

the term (x-5) now appears on both parts of the equation and can be factored as follows:

`(x^2 +6)*(x-5)=0`

From here, we can find all three roots to the function:

`(x-5)=0\\x_(1) =5\\x^2+6=0\\x=√(-6)`

The only real root is 5 since there are no real square roots for negative numbers, the complex roots are:

`x=√(6)√(-1)\\x_(2)=i√(6)\\x_(3)=-i√(6)\\`

The zeroes of the polynomial funtion are:

`x_(1) =5\\x_(2)=i√(6)\\x_(3)=-i√(6)\\`