To factor a trinomial in the form x2+bx+c, you have only to remember that:
x2+bx+c=(x+a)(x+b)=x2+ax+bx+ab=x2+(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 x2 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 x2+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 x2+12x+20 are (x+2) and (x+10):
x2+12x+20=(x+2)(x+10)
Factor m2-7m+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 m2-7m+10 are (x-2) and (x-5):
m2-7m+10=(x-2)(x-5)
