Question 19916
{{{2c^3-5c^2+c+2}}}


By trial and error, put values into the equation. This will lead us to find one of the factors... start with easy numbers like +1, -1 then +2 and -2 etc until you find one.


so, if c=1, we get that the polynomial is equal to zero, so c=1 is a solution (or root). If c=1, then (c-1)=0. Therefore, we can say that (c-1) is a factor.


You now need to perform long division of this factor into the polynomial to find the other factor. Doing this, gives {{{2c^2-3c-2}}}.


So, the factors are {{{(c-1)(2c^2-3c-2)}}}. Factoring the quadratic gives the final answer: {{{(c-1)(2c+1)(c-2)}}}


jon.