To factor a trinomial in the form x^{2}+bx+c, you have only to remember that:

x^{2}+bx+c=(x+a)(x+b)=x^{2}+ax+bx+ab=x^{2}+(a+b)x+ab

The coefficient of the middle term is the sum of a and b.

The last term is the product of a and b.

Therefore, to factor a trinomial in which the coefficient of x^{2} is 1, we need only find the numbers a and b whose sum is the coefficient of the middle term and whose product is the constant term (last term).

Factor x^{2}+12x+20

We need two numbers whose sum is 12 and whose product is 20.

Here are all the possibilities for products that are 20:

Product |
Sum |

1·20=20 |
1+20=21 |

2·10=20 |
2+10=12 |

4·5=20 |
4+5=9 |

The second line gives us what we want.

The factors of x^{2}+12x+20 are (x+2) and (x+10):

x^{2}+12x+20=(x+2)(x+10)

Factor m^{2}-7m+10

We need two numbers whose sum is -7 and whose product is 10. Therefore, we are looking for two negative numbers.

Here are all the possibilities for products that are 10:

Product |
Sum |

(-1)·(-10)=10 |
-1-10=-11 |

(-2)·(-5)=10 |
-2-5=-7 |

The second line gives us what we want.

The factors of m^{2}-7m+10 are (x-2) and (x-5):

m^{2}-7m+10=(x-2)(x-5)