## Factoring Trinomials

Specifically, factoring trinomials *where a is greater than one* (assuming the form *ax*^{2}+*bx*+*c*).

Everyone hates these. You can’t factor them using the method normally taught in school, so many teachers just teach “Guess and Check” for this type of trinomial, in the hopes that they’ll get good enough at guessing and checking.

This is dumb.

There IS a method to factor this sort of trinomials; it’s called The British Method. It’s the only memorable thing my Algebra II teacher taught me. Ready? Here we go.

Here it is in its generality, i.e. with variables | And here it is as an example. |

1. Realize that your trinomial is of the form ax^{2}+bx+c |
1. 2x^{2}+x-1 |

2. Find a number s such that a*c=s. |
2. 2*-1=-2. |

3. Find two numbers p and q such that p*q=s and p+q=b. |
3. p=2; q=-1: 2*-1=-2 and 2+-1=1. |

4. Split the middle term into two binomials, thus: ax^{2}+px and qx+c |
4. 2x^{2}+2x and -x-1 |

5. Factor completely the left binomial to obtain kx(ex+f). |
5. 2x(x+1) |

6. Factor the right binomial such that you get m(ex+f). |
6. -(x+1) |

7. Create two binomials (kx+m) and (ex+f); these are your factors! |
7. (2x-1)(x+1) |

That technique might be good to get even more intuition however we’re all taught the technique of completing the square, thus:

2x^2 + x -1

2(x^2 +(1/2)x – (1/2))

2((x+1/4)^2 – 1/16 – 1/2)

2((x+1/4)^2 – 9/16)

2(x+1/4+3/4)(x+1/4-3/4)

2(x+1)(x-1/2)

which will work every time with quadratic polynomials.

Edit: the page didn’t parse my reply correctly

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