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Factor a formula of form ax^2+bx by pulling out an x. This is called "factoring out a common factor."

For example, 3x^2+5x can be split up by pulling out the x and getting x*(3x+5), where the asterisk * indicates multiplication.

  • 2

Factor a formula of form ax^2+bxy+cy^2+d--i.e., a form with at least two variables and at least four terms--by grouping.

For example, take 6x^3-9x^2+2xy-3y. Group terms that have variables in common: (6x^3-9x^2)+(2xy-3y). Pull out common terms: 3x^2(2x-3)+y(2x-3). Group again: (3x^2+y)(2x-3). This method doesn't always work. Usually the problem states to use grouping so that you don't waste a lot of time on a method that won't work.

  • 3

Factor a formula of form ax^2+bx+c (with a, b, c constants) by grouping.

For example, for 6x^2+17x+12, we have a=6, b=17 and c=12. Use the following algorithm to factor it: find factors of ac that add to b; ac is 72. The first factorization that comes to most people's minds is 9*8. Fortunately, 9+8=17. Now rewrite the equation with b split up into 9 and 8, and then factor by grouping the four-term polynomial: 6x^2+(9+8)x+12 = 6x^2+9x+8x+12 = (6x^2+9x)+(8x+12) = 3x(2x+3)+4(2x+3). Group again to get (3x+4)(2x+3).

  • 4

Refer to formulas when it helps. For example, the formula for x^2-y^2 is (x-y)(x+y), so, for instance, 9x^2-4=(3x-2)(3x+2).

The formula for factoring x^3-y^3 is (x-y)(x^2+xy+y^2). The formula for x^3+y^3 is (x+y)(x^2-xy+y^2).

  • 5

Use two substitutions to solve cubics (single-variable polynomials of order 3), a method first published by Gerolamo Cardano. For the form x^3+ax^2+bx+c=0, substitute x=z-a/3, where z is a variable. Then regroup terms of the same order. (3b-a^2)/3 will end up being the coefficient of z. Then define p=(3b-a^2)/3. Make the substitution z=w-p/3x (called "Vieta's substitution"). Then regroup terms of common order again. The resulting formula is a quadratic in w^3. If the formula can't be factored using grouping as described above, factor quadratics using the quadratic equation.



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