How do I factor the the following cubic polynomial: $x^3-12x^2+41x-42$?

Question

$$x^3-12x^2+41x-42$$

What is the most efficient/easy way to do this?

Answer 1

hint

Observe that

$$x^3-12x^2+41x-42=$$

$$(x-4)^3-7(x-4)-6$$

Now, check the divisors of $ 6 $ which are roots of

$$y^3-7y-6$$

You will see that $ -1,-2$ and $3$ work.

the roots of the initial equation are then $-1+4=3; \; -2+4=2 $ and $3+4=7$.

Without divisors, we have $$y^3-7y-6=(y^3+1)-7(y+1)$$ $$=(y+1)(y^2-y+1-7)$$ $$=(y+1)(y+2)(y-3)$$

Answer 2

Using Vieta's is really easy.

The sum of the roots is $-(-12) = 12$, and the product of the roots is $-(-42) = 42$.

Find three numbers that sum to $12$ and multiply to $42$. You should get $2, 3, \text{and } 7$.

Therefore, the factorization is $$(x-2)(x-3)(x-7)$$