Factoring ax2 + bx + c
If a = 1
Suppose b and c are fixed numbers and that x can vary. Factoring x2 + bx + c means finding two
numbers n and m such that x2 + bx + c = (x + n)(x + m) for all x. To find the numbers n and m, use
the fact that x2 + bx + c = (x + n)(x + m) = x2 + (n + m)x + nm. By equating these trinomials and
comparing the coefficients of x, we see that b = (n + m) and c = nm.
So here is a method that these facts suggest. Let’s illustrate it by factoring x2 − 7x + 12. First, find
all the prime factors of 12. 12 = 2 · 2 · 3. Next, create two columns labeled b and c. Start with the
right-hand column for c and write down all the different ways that c is the product of two numbers.
Next, in column b, write down all the sums from the two factors in column c. You’ll notice that
c = −3 · −4 = 12 and b = −3 − 4 = −7. So we have found n and m. n = −3 and m = −4. The full
answer then is x2 − 7x + 12 = (x − 3)(x − 4). Here is the work:
x2 − 7x + 12
b = (n + m)
13
8
7
-13
-8
-7
c = nm
1 · 12
2·6
3·4
-1 · -12
-2 · -6
-3 · -4
Once you factor enough trinomials of the form x2 + bx + c = (x + n)(x + m), you will have developed
the intuition to only have to check a few possible combinations of n and m.
If a 6= 1
Factoring ax2 +bx+c means finding four numbers h, n, k, m such that ax2 +bx+c = (hx+n)(kx+m).
Checking all the possible combinations of h, n, k, m could be frustrating. Here is a nice trick to finding
these numbers.
To factor ax2 + bx + c in the case where a 6= 1, first factor the simpler trinomial x2 + bx + ac. Then,
with your newly discovered n and m, write x2 + bx + ac = (x + n)(x + m).
n m
The next step is to write ax2 + bx + c = a x +
x+
. That is a correct equation.
a
a
The last step is to beautify the right-hand side of this equation if you can.
Consider this example
Factor 2x2 − 7x + 6. The trick is to first factor x2 − 7x + 2 · 6. We already know the answer to that.
x2 − 7x + 12 = (x − 3)(x − 4).
The next step is to remember the 3 places to insert the a and write 2x2 − 7x + 6 = 2(x − 3/2)(x − 4/2).
And the last step only requires cleaning up the right-hand side. 2x2 − 7x + 6 = (2x − 3)(x − 2).
The reason this method works
For any numbers a, b, c, we have found two other numbers n and m such that n + m = b and nm = ac.
n=
b+
√
b2 − 4ac
b−
and m =
2
√
b2 − 4ac
.
2
Consequently (x + n)(x + m) = x2 + (n + m)x + nm = x2 + bx + ac.
n m
n+m
nm
nm
2
Similarly a x +
x+
=a x +
x + 2 = ax2 + (n + m)x +
= ax2 + bx + c.
a
a
a
a
a
everythingimportant.org/trinomials.pdf