Quadratic Factoring
Suppose we want to factor 6x^2+11x-10. We can use trial and error, go through every combination, and eventually get to a solution, but we might unknowingly skip possible pairs. Surely there is a systematic way to do this. Here it is:
We want to split the 11x into ax+bx such that there is a common ratio 6:a and b:-10
In other words, we want to rewrite 6x^2+11x-10 as 6x^2+ax+bx-10
such that 6/a = b/-10 and such that a+b=11.
How do we find such an a and b?
Notice that if we cross-multiply, 6/a = b/-10 is the same as ab = -60
So we need a and b such that ab = -60 and a+b=11; in other words, two numbers that add to 11 and multiply to -60.
Let's start with the "multiply to -60" part. We will list the factors in order, then consider their sum:
1 -60 sum = -59
2 -30 sum = -28
3 -20 sum = -17
4 -15 sum = -11
Hold it right there. 4 and -15 add to -11, so if we did -4 and 15, the sum would be 11.
So a = -4 and b = 15.
Now we rewrite 6x^2+11x-10 as 6x^2 - 4x + 15x - 10, which we can now factor by grouping:
(6x^2 - 4x) + (15x - 10) =
2x(3x-2) + 5(3x-2) =
(3x-2)(2x+5)
So 6x^2+11x-10 factors as (3x-2)(2x+5):
6x^2+11x-10 = (3x-2)(2x+5)
For this specific example, it actually may have been faster to just use trial and error, but for more complicated factorizations with many possibilities, trial and error is going to take forever.
Try to factor 36x^2-73x-18 with trial and error, for example.
The short way of saying this is that if you want to factor ax^2+bx+c, we need two numbers that multiply and equal ac and that add to equal b.
We want to split the 11x into ax+bx such that there is a common ratio 6:a and b:-10
In other words, we want to rewrite 6x^2+11x-10 as 6x^2+ax+bx-10
such that 6/a = b/-10 and such that a+b=11.
How do we find such an a and b?
Notice that if we cross-multiply, 6/a = b/-10 is the same as ab = -60
So we need a and b such that ab = -60 and a+b=11; in other words, two numbers that add to 11 and multiply to -60.
Let's start with the "multiply to -60" part. We will list the factors in order, then consider their sum:
1 -60 sum = -59
2 -30 sum = -28
3 -20 sum = -17
4 -15 sum = -11
Hold it right there. 4 and -15 add to -11, so if we did -4 and 15, the sum would be 11.
So a = -4 and b = 15.
Now we rewrite 6x^2+11x-10 as 6x^2 - 4x + 15x - 10, which we can now factor by grouping:
(6x^2 - 4x) + (15x - 10) =
2x(3x-2) + 5(3x-2) =
(3x-2)(2x+5)
So 6x^2+11x-10 factors as (3x-2)(2x+5):
6x^2+11x-10 = (3x-2)(2x+5)
For this specific example, it actually may have been faster to just use trial and error, but for more complicated factorizations with many possibilities, trial and error is going to take forever.
Try to factor 36x^2-73x-18 with trial and error, for example.
The short way of saying this is that if you want to factor ax^2+bx+c, we need two numbers that multiply and equal ac and that add to equal b.
