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**HallsofIvy** Okay, you can factor here as well. The only way to factor "2x^2", with integer coefficients, is (2x)(x). The only ways to factor "-3" are either (3)(-1) or (-3)(1). So we know we must have one of (2x+3)(x- 1) or (2x- 3)(x+ 1) or (2x+ 1)(x- 3) or (2x-1)(x+ 3). Testing each of those:

(2x+ 3)(x- 1)= 2x^2+ 3x- 2x- 3= 2x^2+ x- 3.

(2x- 3)(x+ 1)= 2x^2- 3x+ 2x- 3= 2x^2- x- 3.

(2x+1)(x- 3)= 2x^2+ x- 3x- 3= 2x^2- 2x- 3.

(2x- 1)(x+ 3)= 2x^2- x+ 3x- 3= 2x^2+ 2x- 3.

NONE of those is equal to 2x^2- 5x- 3. It **cannot** factored with integer coefficients so it **cannot** reduce to (x- 3)/(3x^2+ 1)