Find the values of b and c for which x^2+bx+c is a factor of 2x^3-3x^2-3x-5
Ok, i've tried to find a factor of the quadratic function using trial and error.
Let f(x) = 2x^3-3x^2-3x-5
f(1) is not equals to 0
f(-1) is not equals to 0
...
and i cant find a number that would equate the function to 0. So i'm stuck and i don't know how i should go about doing this question. Help!
Use long division: divide 2x^3-3x^2-3x-5 by x^2+bx+c
You want the remainder to be 0, so you should be able to solve for b and c after you divide.