# Algebra: Factoring with the Bomb Method

## Factoring with the Bomb Method

I'll be honest; no one but me calls this little factoring technique "the bomb." In fact, I know of few people that teach it at all, because when it comes to factoring tricky quadratic trinomials (tricky because their leading coefficients aren't 1), most algebra teachers tell you to "play around" with binomials until you find something that works.

That is a startling lack of direction in the face of truly tough factoring problems, especially considering that just about everything else you learn in algebra has 1,241,933 steps you have to follow in *exactly* the right order, or else the ghosts of long-dead mathematicians rise from their graves, kick you in the shins, and mark points off your test.

I learned this method many moons ago, under the name "factoring by decomposition," but that name sounds more like a term tossed around at a mortuary (speaking of dead mathematicians). It deserves a name that says "I am an all-but-forgotten method that will help you factor any *nonprime* trinomial in the known universe," so I decided to rename it.

Here are the steps that comprise the bomb technique of factoring the polynomial *ax*^{2} + *bx* + *c*:

##### Talk the Talk

Like a number, if a polynomial cannot be factored, it is considered **prime**.

**Factor out the GCF, if it exists**. This should always be your first step in every factoring problem.**Find two mystery numbers**. These are similar to the numbers you sought when factoring simpler trinomials, but just a bit different. You still want the sum to be the x-coefficient (*b*), but now you want their product to equal*ac*, the product of the leading coefficient and the constant.**Replace the**. Rewrite the polynomial, but where*x*-coefficient*b*once stood, write the sum of the two mystery numbers in parentheses. This powerful little clump of parentheses is about to blow this trinomial up into two binomials; that's why I call this method the bomb.**Distribute**. There's still an*x*through the parentheses*x*multiplied times the quantity in parentheses from step 3; multiply each of the mystery numbers by*x*to eliminate those parentheses.**Factor by grouping**. The end result, as it was before when you factored by grouping, will be the product of two binomials.

The best way to get really good at this method is to practice a lot. Make up your own practice trinomials to factor by multiplying simple binomials together; and then try to factor the result back into those binomials.

**Example 5**: Factor the polynomial 6*x*^{2} - *x* - 12.

**Solution**: This polynomial has no GCF, so you skip right to calculating the mystery numbers; they should add up to -1 and have a product of 6(-12) = - 72. Those numbers, then, must be -9 and 8. Replace the *x* coefficient of -1 with the sum of those mystery numbers in parentheses.

- 6
*x*^{2}+ (-9 + 8)*x*- 12

##### You've Got Problems

Problem 5: Factor the polynomial 4*x*^{2} + 23*x* - 6.

Distribute the *x* lying outside the parentheses.

- 6
*x*^{2}- 9*x*+ 8*x*- 12

Now you can factor by grouping to finish.

- 3
*x*(2*x*- 3) + 4(2*x*- 3) - =(2
*x*- 3)(3*x*+ 4)

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

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