Factoring Trinomials by Grouping
Factoring a Trinomial of the Form ax^{2} + bx + c by Grouping
Example 1Factor: 32x^{2}  20x  3
Solution
Step 1 Factor out common factors (other than 1 or 1).
There are no common factors other than 1 and 1.
Step 2 List the values of a, b, and c. Then find two integers whose
product is ac and whose sum is b.
32x^{2}  20x  3 has the form ax^{2} + bx + c where a = 32, b
= 20, and c = 3.
The product ac is 32 Â· (3) = 96. Thus, find two integers whose
product, ac, is 96 and whose sum, b, is 20.
â€¢ Since their product is negative, the integers must have different signs.
â€¢ Also, their sum is negative, so the integer with the greater absolute
value must be negative.
Here are some of the possibilities:
Product
1 Â· (96)
2 Â· (48)
3 Â· (32)
4 Â· (24) 
Sum 95
46
29
20 
Since 4 and 24 have product 96 and sum 20, we do not need to
consider any other pairs of integers.
Step 3 Replace the middle term, bx,
with a sum or difference using the
two integers found in Step 2. 
32x^{2}  20x  3

Replace 20x with 4x  24x.
Step 4 Factor by grouping.
Group the first pair of terms and group
the second pair of terms.
Factor 4x out of the first group; factor
3 out of the second group.
Factor out the common factor, (8x + 1). 
= 32x^{2} + 4x  24x  3
= (32x^{2} + 4x) + (24x  3)
= 4x(8x + 1) + (3)(8x + 1)
= (8x + 1)(4x  3) 
The result is: 32x^{2}  20x  3
= (8x + 1)(4x  3)
Note:
We replaced 20x with 4x  24x. If we
switch 4x and 24x, we can still group
and factor:
= 32x^{2}  24x + 4x  3
= (32x^{2}  24x) + (4x  3)
= 8x(4x  3) + 1(4x  3)
= (4x  3)(8x + 1)
Example 2
Factor: 2x^{2} + 4x + 3.
Solution
Step 1 Factor out common factors (other than 1 or 1).
There are no common factors other than 1 and 1.
Step 2 List the values of a, b, and c. Then find two integers whose
product is ac and whose sum is b.
2x^{2} + 4x + 3 has the form ax^{2} + bx + c where a = 2, b = 4, and c
= 3.
The product ac is 2 Â· 3 = 6.
Thus, find two integers whose product, ac, is 6 and whose sum, b, is 4.
â€¢ Since their product is positive, the integers must have the same sign.
â€¢ Since their sum is also positive, the integers must both be positive.
Here are the possibilities:
Product
1 Â· (6)
2 Â· (3) 
Sum 7
5 
Neither possibility has the required sum, 4.
Since there are no two integers whose product is 6 and whose sum is 4, we
conclude that 2x^{2} + 4x + 3 is not factorable over the integers.
Note:
This approach tells us directly when the
trinomial is not factorable.
Thatâ€™s a major advantage of this method.
Example 3
Factor: 2x^{2}  8x  10
Solution
Step 1 Factor out common factors (other than 1 or 1).
Factor out the common factor of 2.
The trinomial has the form x^{2} + bx + c. 
2x^{2}  8x  10
= 2(x^{2}  4x  5) 
Since the coefficient of the x^{2}term is 1, we can factor the trinomial by
the productsum method. 
That is, we find two integers whose
product is 5 and whose sum is 4.
The integers are 5 and 1. 
= 2(x  5)(x + 1) 
The result is: 2x^{2}  8x  10 = 2(x  5)(x + 1).
You can multiply to check the factorization. We leave the check to you.
