If the quadratic expression ( 2 x^2 + 11 x - 21) factors, the possible integer-based factorings would be (2x - 21 ) ( x + 1) (2x + 21 ) ( x - 1) (2x + )(x - 21) (2x - )(x + 21 ) (2x - 3 ) ( x + 7 ) (2x + 3 ) ( x - 7 ) (x - 3 ) ( 2x + 7 ) (x + 3) ( 2x - 7 ). When multiplied out each of these expressions will give you the required 2 x^2 and the -2. However at most one of them will give you the + 11 x you need. For example (2x - 21 ) ( x + ) = 2x ( x + ) - 21 ( x + ) = 2 x^2 + 2 x - 21 x - 21 = 2 x^2 - 19 x - 21. You get -19 x , not +11 x . Check out the rest of the possibilities. I do believe that one of them works. ***************** A slightly more complicated example involving both x and y and factoring a monomial: Factor into linear factors 6x^3 + 33x^2y - 63y^2x Start by factoring out x to get x(6x^2+33xy-63y^2) Then factor the trinomial 6 x^2 + 33 x y - 63 y^2: If you had factored out 3x you would have obtained 3x ( 2 x^2 + 11 x y - 21 y^2). If the quadratic expression ( 2 x^2 + 11 x y - 21 y^2) factors, the possible integer-based factorings would be (2x - 21 y) ( x + y) (2x + 21 y) ( x - y) (2x + y)(x - 21y) (2x - y)(x + 21 y) (2x - 3 y) ( x + 7 y) (2x + 3 y) ( x - 7 y) (x - 3 y) ( 2x + 7 y) (x + 3y) ( 2x - 7 y). When multiplied out each of these expressions will give you the required 2 x^2 and the -21 y^2. However at most one of them will give you the + 11 x y you need. For example (2x - 21 y) ( x + y) = 2x ( x + y) - 21 y ( x + y) = 2 x^2 + 2 x y - 21 x y - 21 y^2 = 2 x^2 - 19 x y - 21 y^2. You get -19 x y, not +11 x y. Check out the rest of the possibilities. I do believe that one of them works.