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November 28, 2014

November 28, 2014

Posted by **Marrion** on Saturday, September 26, 2009 at 2:53am.

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 dont know how i should go about doing this question. Help!

- ?? Maths?? -
**Marth**, Saturday, September 26, 2009 at 10:12amUse 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.

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