If one root of a cubic is known (and verified), then we can do a synthetic division and factor the remaining quadratic expression.
f(x)=x^3+x^2-22x-40
f(5)=125+25-22(5)-40=0 so x-5 is a factor
Now do the synthetic division
5 | +1 +1 -22 -40
--------------------
1 6 8 0
The resulting quadratic is x^2+6x+8 which can be factored into
(x+2)(x+4)
The complete factorization is therefore
f(x)=x³ + x² - 22x - 40 = (x-5)(x+2)(x+4)
