Question

X^3 - 3x^2 + 5x - 15 = 0

What real value of x is a solution to the above equation?

Answer 1

It is often difficult to factor a polynomial, but notice that the ratio of the coefficient of the first term to that of the second term is 1:3, and that the ratio of the coefficient of the third term to that of the fourth is also 1:3. Therefore, we can factor this cubic by grouping.


`x^3-3x^2+5x-15=0\\x^2(x-3)+5(x-3)=0\\(x^2+5)(x-3)=0\\x^2+5=0,\ x-3=0`

We have three solutions in all.

Two imaginary:

`x^2=-5\\x=\pm i √(5)`

One real:

`x=3`