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Factoring Quadratic Polynomials
In this handout we will factor quadratic polynomials of the form ax^{2} + 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(6x^{2}  3x + 2x  1) = 3(6x^{2}  x  1) = 18x^{2}  3x
 3 .
(b) Factor 18x^{2}  3x  3 .
Solution: 18x^{2}  3x  3 = 3(6x^{2}  x  1) = 3(6x^{2}  3x + 2x  1) = 3(3x + 1)(2x 
1) .
Example 2. (a) Multiply (5x + 1)(5x  1) .
Solution: (5x + 1)(5x  1) = 25x^{2}  5x + 5x  1 = 25x^{2}  1 .
(b) Factor 25x^{2}  1 .
Solution: 25x^{2}  1 = 25x^{2}  5x + 5x  1 = (5x + 1)(5x  1) .
Notice that the last step in factoring was simply factoring by (re)grouping:
In Example 1: 6x^{2}  3x + 2x  1 = 3x(2x  1) + 1(2x  1) = (3x + 1)(2x  1) ;
and
in Example 2: 25x^{2}  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) = mpx^{2} + mqx + npx + nq = mpx^{2} + (mq + np)x + nq = ax^{2} + 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 ax^{2} + 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 4x^{2} + 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:
First factor:  1  2  4  7  14  28 
Second factor:  28  14  7  4  2  1 
Their sum:  27  12  3  3  12  27 
The fourth column tells us to try 3x = 7x  4x . We have 4x^{2} + 3x  7 = 4x^{2} +
7x  4x  7
= x(4x + 7)  1(4x + 7) = (4x + 7)(x  1) .
Example 4. Factor 49x^{2}  1 .
Solution: Here ac = 49(1) = 49 , b = 0
First factor:  1  7 . . . 
Second factor:  49  7 . . . 
Their sum:  48  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 49x^{2}  1 = 49x^{2}  7x + 7x  1 =
7x(7x  1)
+ (7x  1) = (7x  1)(7x + 1) .
Example 5. Factor 4x^{2}  4x + 1 .
Solution: Here ac = 4 × 1 = 4 , b = 4 . The (relevant part of the) table is now
First factor:  1  2 . . . 
Second factor:  4  2 . . . 
Their sum:  5  4 . . . 
Replacing 4x by 2x  2x we have 4x^{2}  4x + 1 = 4x^{2}  2x  2x + 1 = 2x(2x 
1)  (2x  1)
= (2x  1)(2x  1) = (2x  1)^{2} .
Example 6. Factor 5x^{2} + 7x + 3 .
Solution: Here ac = 5 × 3 = 15 , b = 7 . The table is
First factor:  1  3  1  3 
Second factor:  15  5  15  5 
Their sum:  16  8  16  8 
In this example the sum is never equal to 7 . Since we have tried every
possible factoring of 15 ,
the conclusion is that 5x^{2} + 7x + 3 is not factorable over the integers.
Example 7. Factor 6x^{2} + 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
First factor:  1  2  1  2 
Second factor:  6  3  6  3 
Their sum:  7  5  7  5 
Using 5xy = 2xy + 3xy (from the last column) we have 6x^{2} + 5xy + y2 = 6x^{2} +
2xy + 3xy + y2
= 2x(3x + y) + y(3x + y) = (3x + y)(2x + y) .
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