Factoring Quadratic Polynomials

In this handout we will factor quadratic polynomials of the form ax² + bx + c , where a ,
b , and c are integers and a≠0 . Recall that factoring is actually the reverse process of multiplying,
for example:

Example 1. (a) Multiply 3(3x + 1)(2x - 1) .
Solution : 3(3x + 1)(2x - 1) = 3(6x² - 3x + 2x - 1) = 3(6x² - x - 1) = 18x² - 3x - 3 .
(b) Factor 18x² - 3x - 3 .
Solution: 18x² - 3x - 3 = 3(6x² - x - 1) = 3(6x² - 3x + 2x - 1) = 3(3x + 1)(2x - 1) .

Example 2. (a) Multiply (5x + 1)(5x - 1) .
Solution: (5x + 1)(5x - 1) = 25x² - 5x + 5x - 1 = 25x² - 1 .
(b) Factor 25x² - 1 .
Solution: 25x² - 1 = 25x² - 5x + 5x - 1 = (5x + 1)(5x - 1) .
Notice that the last step in factoring was simply factoring by (re)grouping:

In Example 1: 6x² - 3x + 2x - 1 = 3x(2x - 1) + 1(2x - 1) = (3x + 1)(2x - 1) ; and
in Example 2: 25x² - 5x + 5x - 1 = 5x(5x - 1) + 1(5x - 1) = (5x + 1)(5x - 1) .

It remains to explain how to regroup or split the middle term bx ; in 1.b we replaced -x by
-3x + 2x , and in 2.b we replaced 0x by -5x + 5x . How do we determine those coefficients in
more complicated cases? Let us consider the general case.

Suppose that it is possible to write the given polynomial ax ² + bx + c as a product of the
form (mx + n)(px + q) where m , n , p , and q are integers to be determined. Then we have

(mx + n)(px + q) = mpx² + mqx + npx + nq = mpx² + (mq + np)x + nq = ax² + bx + c ,
where mp = a , mq + np = b , nq = c . Written in reverse, this becomes an exercise in factoring;
the only problem is to replace b by the sum mq + np . We already know that the sum of mq
and np must be equal to b ; their product is (mq)(np) = (mp)(nq) = a × c . In other words, we
are looking for two integers whose sum is b and whose product is ac . The factoring method for
polynomials of the form ax² + bx + c can now be stated as follows:

(1) Calculate ac . Find two integers whose product is ac and whose sum is b .
(2) Replace the middle term, bx , by the sum of two terms whose coefficients are integers
found in (1).
(3) Factor by (re) grouping .

Notice that if ac ≠ 0 there are only finitely many ways to represent ac as a product of
two integers , so that the method is guaranteed to produce a factoring or to show that no factoring
(over the integers) is possible. If ac = 0 just factor directly or use b = 0 + b .

Example 3. Factor 4x² + 3x - 7 .

Solution: Here ac = 4(-7) = -28 , b = 3 , so we are looking for two numbers whose
product is -28 and whose sum is 3 . It is convenient to put the possible factors in a table:

The fourth column tells us to try 3x = 7x - 4x . We have 4x² + 3x - 7 = 4x² + 7x - 4x - 7
= x(4x + 7) - 1(4x + 7) = (4x + 7)(x - 1) .

Example 4. Factor 49x² - 1 .
Solution: Here ac = 49(-1) = -49 , b = 0

The second column is the good one (the sum is zero as required ), so we omit the rest of the table.
This means 0x = -7x + 7x should work. We have 49x² - 1 = 49x² - 7x + 7x - 1 = 7x(7x - 1)
+ (7x - 1) = (7x - 1)(7x + 1) .

Example 5. Factor 4x² - 4x + 1 .

Solution: Here ac = 4 × 1 = 4 , b = -4 . The (relevant part of the) table is now

Replacing -4x by -2x - 2x we have 4x² - 4x + 1 = 4x² - 2x - 2x + 1 = 2x(2x - 1) - (2x - 1)
= (2x - 1)(2x - 1) = (2x - 1)² .

Example 6. Factor 5x² + 7x + 3 .
Solution: Here ac = 5 × 3 = 15 , b = 7 . The table is

In this example the sum is never equal to 7 . Since we have tried every possible factoring of 15 ,
the conclusion is that 5x² + 7x + 3 is not factorable over the integers.

Example 7. Factor 6x² + 5xy + y2 .
Solution: Although this trinomial contains two variables, the same method still works.
Here we have ac = 6 × 1 = 6 , b = 5 and the table is

Using 5xy = 2xy + 3xy (from the last column) we have 6x² + 5xy + y2 = 6x² + 2xy + 3xy + y2
= 2x(3x + y) + y(3x + y) = (3x + y)(2x + y) .