Solving for a cubic polynomial's roots using Viete's Theorem

Question

I am asked to find the roots $f(x)=x^3-24x^2-24x-25$. However, the only thing I am aware of in regards to finding the roots of cubic polynomials is Viete's Theorem. However, this theorem requires that the polynomial be in the form $f(x)=x^3+qx+r$ i.e. without the quadratic element. Am I missing how Viete can be used to answer this question?

Answer 1

In order to set the quadratic coefficient with zero, replace $x$ with $x+8$. This will shift all of the roots by 8 and the resulting polynomial (once you expand out the powers of $x+8$) will be of the desired form.

In general, to convert a cubic of form $x^3 + ax^2+bx+c$ to one of form $x^3 + qx + r$, you replace $x$ with $x-a/3$.

E: For this example in specific, it might be more effective to first look for rational roots, as mentioned by lhf above in comments. While careful use of the cubic formula will give you a real root, it might be in an ugly form that's not obvious how to simplify.