You may also use the following method to factorize the quadratic expression, such that:

`2x^2 - 5x - 3 = 2(x - x_1)(x - x_2)`

`x_(1,2)` represent the solution to equation `2x^2 - 5x - 3 = 0`

You should use quadratic formula to evaluate `x_(1,2),` such that:

`x_(1,2) = (5+-sqrt(25 + 24))/4 => x_(1,2) = (5+-7)/4`

`x_1 = 3; x_2 = -1/2`

`2x^2 - 5x - 3 = 2(x - 3)(x - (-1/2))`

`2x^2 - 5x - 3 = (x - 3)(2x + 1)`

**Hence, evaluating the factored form of quadratic expression, using the alternative method, yields **`2x^2 - 5x - 3 = (x - 3)(2x + 1).`

`2x^2-5x-3`

`= 2x^2-6x+x-3`

`= 2x(x-3)+1(x-3)`

`= (x-3)(2x+1)`

So when you factorize `2x^2-5x-3 ` you will get;

`2x^2-5x-3= (x-3)(2x+1)`

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