How do you factor the expression #x^3-2x^2-9x+18#?

1 Answer
Jan 19, 2016

#f(x)= (x-2)(x^2-9)# = (x-2)(x-3)(x+3)


We can factorise by determining zeros of f(x). These zeros are some factors of 18. These can be determined by hit and trial method. If at least one zero is located, then remaining two can be used by factorising a quadratic.

In the present case it can be easily verified that x=2 is a zero of the polynomial. Now divide the polynomial by x-2 using long or synthetic division, the quotient got would be #x^2-9#

#f(x)= (x-2)(x^2-9)# = (x-2)(x-3)(x+3)