I discovered something interesting about factoring polynomials that can really speed up the process if you’re like me and the answer doesn’t always just jump out at you.
It’s based on the realization that if one of your polynomial factors can be reduced, then so can your polynomial. Here’s why:
Assume your factors are (akx + bk) (cx + d) (the first factor can be reduced by k)
If you multiply them, you get: akcx2 + akdx + bkcx + bkd, which can be reduced by the same k to
k*(acx2 + adk + bcx + bd)
This is a powerful realization because it means that once you’ve reduced your polynomial, you can automatically eliminate any possible factors if either of them can be reduced.
Consider 6x2 – 7x – 20
Let (ax +b) and (cx + d) be the two factors
Because the x2 coefficient is 6, we know that a * c = 6. There are four possible ways for this to occur:
1*6
2*3
3*2
6*1
Because the magnitude of the purely numeric value is 20, we know that b * d = 20 (ignoring the sign for now). There are six possible ways for this to occur:
1*20
2*10
4*5
5*4
10*2
20*1
Each combination of the first can go with any combination of the second, which leaves us with 4 * 6 possibilities. However, the last half are identical to the first half, with just the order changed, so we’re left with these 12 possibilities, listed here without commas instead of signs between the x and the numeric terms:
(x, 1) * (6x, 20)
(x, 2) * (6x, 10)
(x, 4) * (6x, 5)
(x, 5) * (6x, 4)
(x, 10) * (6x, 2)
(x, 20) * (6x, 1)
(2x, 1) * (3x, 20)
(2x, 2) * (3x, 10)
(2x, 4) * (3x, 5)
(2x, 5) * (3x, 4)
(2x, 10) * (3x, 2)
(2x, 20) * (3x, 1)
Instead of being forced to multiply all 12 factor pairs to see which gives us the result we want, we can IMMEDIATELY eliminate EIGHT out of TWELVE because one of the factors in them can be reduced: They are
(x, 1) * (6x, 20) right factor can be reduced by 2
(x, 2) * (6x, 10) right factor can be reduced by 2
(x, 5) * (6x, 4) right factor can be reduced by 2
(x, 10) * (6x, 2) right factor can be reduced by 2
(2x, 2) * (3x, 10) left factor can be reduced by 2
(2x, 4) * (3x, 5) left factor can be reduced by 2
(2x, 10) * (3x, 2) left factor can be reduced 2
(2x, 20) * (3x, 1) left factor can be reduced 2
This leaves us with
(x, 4) * (6x, 5)
(x, 20) * (6x, 1)
(2x, 1) * (3x, 20)
(2x, 5) * (3x, 4)
We know each of these combinations will give us the correct coefficients for our x2 and numeric terms, so let’s determine what ‘x’ coefficient each will provide. This is the sum or difference, depending on the sign, of the products of the outer and inner terms.
Because our numeric result is negative (-20) we know the signs of the numeric terms in the factors must be opposites, so our ‘x’ coordinate will be the difference of the inner and outer products. Calculating these is very easy, considering we only have FOUR of TWELVE left.
|
Outer product |
Inner product |
Magnitude of Difference/ ‘x’ coefficient |
|
| (1x, 4) * (6x, 5) |
5 |
24 |
19 |
| (1x, 20) * (6x, 1) |
1 |
120 |
119 |
| (2x, 1) * (3x, 20) |
40 |
3 |
37 |
| (2x, 5) * (3x, 4) |
8 |
15 |
7 |
The only combination that gives us 7 as the x coefficient is the last one so all that’s left to do is determine which term has the negative sign and which has the positive.
(2x + 5) * (3x – 4) = 6x2 -8x + 15x – 20 = 6x2 + 7x – 20
(2x – 5) * (3x + 4) = 6x2 +8x – 15x – 20 = 6x2 – 7x – 20
The second pair contains the correct sign for our ‘x’ coefficient so our factors are:
(2x – 5) * (3x + 4)
To summarize: There is no simple way to factor a polynomial if its x2 term is not 1. If your first few guesses don’t work and you need a systematic approach, you can frequently eliminate more than half the possibilities by first reducing the polynomial so its terms have no common factors and then immediately eliminating all possible factors where either of them can be reduced.
Pretty interesting.
Do you have any other tips for factoring polynomials?