Write $f(x)=x^3+3x^2-25x+21$ in factored form.
The first thing we should do is see if we can factor this polynomial by grouping. That technique won't work on this polynomial. Therefore, we should start looking for roots, because if we find a root, then we find a factor, then we can perform synthetic division to get the original polynomial as a product of a linear factor and a quadratic. From there, we can factor the quadratic using the quadratic formula, or just by factoring it if we can see it's factorization.
The rational roots theorem tells us that the possible rational roots of this polynomial are divisors of the constant term divided by the divisors of the leading term. So the the possible rational roots are $\pm 1, \pm 3, \pm 7, \pm 21$.
I noticed that often times when we have to resort to the possible rational roots, one of the rational roots will be $-1$ or $1$.
$f(-1)= -1 +3 +25 +21 \neq 0$
$f(1) = 1+3-25+21=0$.
So $1$ is a root.
So $(x-1)$ is a factor. We can now do long division or synethic division to get the factorization:
$f(x)=x^3+3x^2-25x+21=(x-1)(x^2+4x-21)$
And we can factor $(x^2+4x-21)=(x+7)(x-3)$
Thus:
$f(x)=x^3+3x^2-25x+21=(x-1)(x^2+4x-21)=(x-1)(x+7)(x-3)$