# At a Glance - Trial and Error

We already know how to factor quadratic polynomials that are the result of multiplying a sum and difference, or the result of squaring a binomial with degree 1. Once in a while, though, trinomials go through mood swings and stop cooperating, and then we have a bit more begging and pleading to do. What do we do in those instances? One method is to try trial and error.

Sounds like something your teacher would advise you not to do, but if you've got a talent for seeing patterns, you like guessing games, you’ve done all your homework and have a lot of time on your hands, or you’re just not a rule follower, this is the method for you. If none of this trial-and-erroring can get a quadratic polynomial out of its bad mood, about all there is left to do is take it for ice cream and then put it down for a nap. Hopefully it won't be quite so pouty when it wakes up.

Remember that a quadratic polynomial is a polynomial of degree 2 of the form *ax*^{2} + *bx* + *c*.

These polynomials are easiest to factor when *a* = 1 (that is, the polynomial looks like *x*^{2} + *bx* + *c*), so we'll look at that case first. Those of you who like torturing yourselves can skip ahead to the harder stuff.

Before we start factoring, we'll revisit multiplication. Assume *m* and *n* are integers. You're not being presumptuous—they *are* integers, we swear. If we multiply:

(*x* + *m*)(*x* + *n*)

...then we find:

*x*^{2} + *mx* + *nx* + *mn*

...which simplifies to:

*x*^{2} + (*m* + *n*)*x* + *mn*

The numbers *m* and *n* multiply to give us the constant term in the final polynomial, and the sum of *m* and *n* is the coefficient of *x* in the final polynomial. Neither *m* nor *n* make an appearance alongside the first term in the final polynomial, which is probably just as well, since that *x* appears to be busy squaring itself.

Let's see how this can be used for factoring by looking at some examples.

### Sample Problem

Factor the polynomial *x*^{2} + 4*x* + 3.

To get a trinomial with an *x*^{2} term, we must have multiplied two binomials, each with an "*x*" term. Wow...it's like we're psychic. By the way, you shouldn't leave your house tomorrow. Don't ask questions.

The original binomials must have looked like this:

(*x* + *m*)(*x* + *n*)

...where *m* and *n* are integers. We need to figure out the values of *m* and *n*. The constant term of the original polynomial is 3, so we need* mn* = 3.

What integers multiply together to give 3? The only choices are 1 and 3, or maybe -1 and -3.

If you can think of any others, congratulations! However, you're wrong. Gee, that victory was short-lived.

The coefficient of the *x* term in the original polynomial is 4, so we also need* m* + *n* = 4.

Since 1 and 3 multiply to give 3 and add together to give 4, we have *m* = 1 and *n* = 3. Therefore, we can factor our original polynomial like this:

*x*^{2} + 4*x* + 3 = (*x* + 1)(*x* + 3)

If we let *m* = 3 and *n* = 1 we'll have the same factorization, except with the factors written in a different order. Either way is correct, so we won't fight about it. We can *all* take turns equaling 3.

### Sample Problem

Factor the polynomial *x*^{2} + 4*x* – 5.

We can factor this quadratic polynomial into two binomials of the form:

(*x* + *m*)(*x* + *n*)

We need to have *mn* = -5. The integers that multiply to give -5 are -1 and 5, or 1 and -5.

We also need to have *m* + *n* = 4, which will limit our options. Not necessarily a bad thing when you're searching for the right answer. The correct choices for *m* and *n* are -1 and 5, and the polynomial factors are:

(*x* – 1)(*x* + 5)

Now that we've gotten some practice with the friendlier varieties of quadratic polynomials, we'll look at general polynomials of the form* ax*^{2} + *bx* + *c *when *a* *doesn't* equal 1.

These guys are of the *un*friendly variety. Don't make them angry. You wouldn't like them when they're angry.

Here's another quick visit to multiplication before we start factoring. The binomials (2*x* + 3) and (*x* + 5) multiply to give us:

2*x*^{2} + 13*x* + 15

The coefficient on the *x*^{2} term is the product of 2 and 1, the coefficients of *x* in each of the binomials:

(**2 x** + 3)(

*+ 5) =*

**x****2**+ 13

*x*^{2}*x*+ 15

The constant term in the product is 3 × 5, the product of the constant terms in the binomials:

(2*x* + **3**)(*x* + **5**) = 2*x*^{2} + 13*x* + 3*x* + **15**

The middle term, 13*x*, is a bit of a mess, but we can make sense of it. Thinking of distribution, 13*x* comes from multiplying the outer terms:

(**2 x** + 3)(

*x*+

**5**) = 2

*x*

^{2}+

**10**+ 3

*x**x*+ 15

...multiplying the inner terms:

(2*x* + **3**)(* x* + 5) = 2

*x*

^{2}+ 10

*x*+

**3**+ 15

*x*...and simplifying (adding the 10*x* and 3*x* together). If we're not greatly mistaken, 10 + 3 = 13. Voila.

Factoring quadratic polynomials involves a bit of trial and error. The more you practice factoring, the less error you'll run into, because you'll learn to see which trials will work without having to write down all the steps. It's like trying to teach yourself to play the piano. You can bang away randomly at the keys for a while, but eventually you'll develop a feel for what note each key is responsible for and your guesswork will become minimized. Hopefully the quadratic polynomials you'll be working on are more "Chopsticks" and less "Flight of the Bumblebee."

### Sample Problem

Factor the polynomial 3*x*^{2} – 2*x* – 1.

If the polynomial factors into two binomials, they'll be of the form:

(*ax* + *b*)(*cx* + *d*)

We need the first numbers in each binomial to have a product of 3, so that means *ac* = 3. We also need the second numbers in each binomial to have a product of -1, so that means *bd *= -1. The only possibilities for *a *and *c* are 3 and 1, since 3 is prime. The last numbers *b *and *d* must be 1 and -1 in order for their product to be -1. With a problem like this, we don't even need to worry about using trial and error. It's more like trial and instant success.

The only question we have left is whether the answer is (3*x* – 1)(*x* + 1) or (3*x* + 1)(*x* – 1).

It's tempting to use "eenie-meenie-miney-mo," but resist the urge. To determine which binomials are the correct factors, we need to figure out which ones will produce the correct *x* coefficient of -2. When we multiply (3*x* – 1)(*x* + 1), here's what we get:

(3*x* – 1)(*x* + 1) = 3*x*^{2} + 2*x* – 1

Well, poop. We want -2*x *in the middle, not 2*x*. Let's try our other option.

(3*x* + 1)(*x* – 1) = 3*x*^{2} – 2*x* – 1

Ah, that's more like it. Consider this puppy factored.

The more you practice factoring, the easier it'll become, and eventually you won't need to keep getting up to sharpen your pencil. In any given example, we can list every single possible factorization...or just the right one. Depends on how long it takes you to find what you're looking for. As long you have the right answer, no one will care if you checked all the possible factorizations. Unless the "All Possible Factorization Monster" truly does exist, but we doubt it. The evidence is shoddy at best.

### Sample Problem

Factor the polynomial -2*x*^{2} + 7*x* – 3.

If this polynomial factors as (*ax* + *b*)(*cx* + *d*), the product of *a *and *c* must be -2, and the product of *b *and *d* must be -3.

So *a* and *c* could be -2 and 1, or 2 and -1.

And *b *and *d* could be -3 and 1, or 3 and -1.

We'll try all the possible factorizations and see which one works. In this case, there are a LOT of possibilities. In fact, it will benefit us to use some factorization organization.

Possible Factorizations |

(-2x – 3)(x + 1) = -2x^{2} – 5x – 3 |

(-2x + 1)(x – 3) = -2x^{2} + 7x – 3 |

(2x – 3)(-x + 1) = -2x^{2} + 5x – 3 |

(2x + 1)(-x – 3) = -2x^{2} – 7x – 3 |

(-2x + 3)(x – 1) = -2x^{2} + 5x – 3 |

(-2x – 1)(x + 3) = -2x^{2} – 7x – 3 |

(2x + 3)(-x – 1) = -2x^{2} – 5x – 3 |

(2x – 1)(-x + 3) = -2x^{2} + 7x – 3 |

From this table, we see that two different factorizations can give us a middle term of 7*x*. What's going on here? What sorcery is this? Okay, let's not be overly dramatic. The fact is that either of these factorizations will work. Watch. We can rewrite (-2*x* + 1)(*x* – 3) by factoring out -1 from the first factor to get:

(-1)(2*x* – 1)(*x* – 3)

Then we can distribute that (-1) back into the second factor to find:

(2*x* – 1)(-*x* + 3)

Either factorization is fine as a final answer. We didn't actually need to check all the possible factorizations, but it's easier to check them all than it is to figure out which ones we could safely ignore. Now we know *exactly* which ones to give the silent treatment to.

Don't worry if trial and error seems a little messy to you. If you like things a bit more clean and organized and all this guessing-and-checking drives you up the wall, we've got another method that works just as well. Read on.

#### Example 1

Factor the polynomial |

#### Example 2

Factor the polynomial 2 |

#### Exercise 1

Factor *x*^{2} + 7*x* – 8.

#### Exercise 2

Factor *x*^{2} + 9*x* + 20.

#### Exercise 3

Factor *x*^{2} – 5*x* – 6.

#### Exercise 4

Factor 3*x*^{2} – 3*x* – 36.

#### Exercise 5

Factor 4*x*^{3} + 44*x*^{2} + 120*x*.

#### Exercise 6

Factor *x*^{2} + 5*x* + 4.

#### Exercise 7

Factor 2*x*^{2} + 13*x* + 15.

#### Exercise 8

Factor 3*x*^{2} + 3*x* – 18.

#### Exercise 9

Factor -6*x*^{2} + 5*x* – 1.

#### Exercise 10

Factor 40*x*^{2} – 46*x* + 12.