A\(2 x^{2} -5x-3=2 x^{2} -6x+x-3\) (I'll explain how I split it in a bit)
\(=2x(x-3)+1(x-3)\) Look for common factors between the 1st two and then the second two while trying to make sure the variables in the brackets are the same.
Now you only have to write the value in the bracket once and then join the outer factors in their own bracket so your final answer will be \((2 x^{2} +1)(x-3)\)
For the first thing I did, I multiplied the first part of the equation by the last part= 2\(x^{2}\) x -3=-6\(x^{2}\). I then proceeded to find two numbers that can be multiplied to give the result and when added will give the middle part (-5x).
Factors of -6\(x^{2}\) are: -1x and 6x, -2x and 3x, -3x and 2x, -6x and 1x. The only ones that add up -5x are -x and 6x so that was why I used them.
b. 3x-2/2x^2-5x-3 - 1/x-3
\(\frac{3x-2}{2 x^{2} -5x-3}- \frac{1}{x-3} = \frac{3x-2}{(2x+1)(x-3)} - \frac{1}{x-3}\)
\(= \frac{1(3x-2) +1(2x+1)}{(2x+1)(x-3)}\)
\(= \frac{3x+2x-2+1}{(2x+1)(x-3) }\)
=\(\frac{x-1}{2 x^{2} -5x-3}\)
