Question 706033
The technique we will use to factor {{{2x^2-5xy-3y^2}}} can be described as "un-FOIL-ing". FOIL is used to multiply expressions like (a+b)(c+d). Since factoring is, in effect, "un-multiplying" an expression we can use the reverse of FOIL to turn {{{2x^2-5xy-3y^2}}} into an expression of the form (a+b)(c+d).<br>
If we have done enough FOIL-ing we know that when (a+b)(c+d) results in a trinomial, like {{{2x^2-5xy-3y^2}}}, then:<ul><li>The first term of the trinomial, {{{2x^2}}}, usually comes from the "F" part of FOIL (which would be a*c). For {{{2x^2}}} there is really only one possible "a" and "c": 2x and x.</li><li>The third term of the trinomial, {{{-3y^2}}}, usually comes from the "L" part of FOIL (which would be c*d). For {{{-3y^2}}} we could have two different c/d pairs: -3y and y or 3y and -y.</li><li>The middle term of the trinomial, -5xy, usually comes from the sum of the "O" and "I" parts of FOIL (which would be a*d + b*c).</li></ul>Now we just try the different possible values for a, b, c and d to see if any of them will create the proper middle term. (If none do then the trinomial will not factor.) As it turns out, there is a combination that works:
a = 2x, b = y, c = x and d = -3y
So {{{2x^2-5xy-3y^2}}} factors into:
(2x + y)(x + (-3y))
or, more simply:
(2x + y)(x - 3y)<br>
To check this, just use FOIL!