Factoring Quadratic Equations
Factoring Quadratic Equations
Factoring Quadratic Equations
Summary
1. Difference of squares .............................................................................................................. 1
2. Mise en évidence simple ......................................................................................................... 2
3. compounded factorization ...................................................................................................... 3
4. Exercises .................................................................................................................................. 7
The goal of this section is to summarize the methods allowing us to factor quadratic
equations, i.e. of form . We will avoid using the famous discriminant
formula
√ 4
2
as much as possible. Three methods allow us to carry out the factoring of most
quadratic functions.
1. Difference of squares
There is a formula that allows for rapid factorization. When a function presents in the
form , it can be factored by the difference of squares formula, i.e.
Hence, you need to obtain the square root of the first and second term and multiply
their sum with their difference. The negative sign separating the terms and is of
capital importance. For example, the rule presented above cannot be applied to
, as the name difference in squares implies!
Example
9 3 3 3
100 10 10 10
81 9 9 9
In the preceding examples, we chose to use values that are perfect squares. This criteria
is not generally necessary. For example, although 8 is not a perfect square, its root is
well defined (it exists) and is equal to √8.
Example
8 √8 √8 √8
3 √3 √3 √3
Also, it is possible that the first term is not but 9 . Once again, you do not need to
worry. We must find the square root of 9 instead of the square root of . In all other
aspects, the method remains the same.
Example
9 25 3 5 3 5 3 5
16 49 4 7 4 7 4 7
8 9 √8 3 √8 3 √8 3
2. Simple factorization
We must avoid confusing the form of the difference of squares x2 ‐ k2 with ax2 ‐ bx. The
presence of an in the second term will allow us to proceed with a simple factorization.
We can bring out the common factor, i.e. the x.
Example
6 6
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Example
Factor 25 10
What are all the factors common to 25 10 ? Each has at least one . The
coefficients 25 and 10 also have the common factor 5. 5 constitutes the greatest
common factor of 25 and 10 . If 5 is removed from these terms (or better
factorized), what remains of 25 and 10 ? Of 25 , 5 will remain. Of 10 , 2 will
remain. Consequently,
25 10 5 5 2
Example
12 12
2 12 2 6
36 6 6 6 1 or‐6 6 1
225 45 15 15 3
9 16 9 16 3 4 3 4
3. Compounded factorization
5 6 2 3 6
Note that we did not cheat : even though we rewrote the function, its total value did
not change. With this new presentation, notice that a factoring can be carried out to the
pairs of terms 2 and 3 6. x is the common factor of 2 , and 3 is
common factor of 3 6. By proceeding to a simple factoring, pair by pair, we obtain
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2 3 6 2 3 2 .
But, this is not all! Now, in this new expression, 2 is a common factor. We can
therefore factor out 2 . Thus, from 2 , x remains. From 3 2 , 3
remains. Consequently,
2 3 2 2 3
Note that we carried out three consecutive factorizations in order to complete the
factoring, as the name compounded factorization implies.
One aspect of the method remains unexplained: why rewrite under the
form 2 3 6 instead of 4 6 or 9
4 6 ? Here is the process allowing us to discover which way to separate the central
term (multiple of x) :
1. the product c
2. the sum
Example
Since 0 and 0, we must first separate the central term 5x into two parts. To do
this, we must find two numbers, m and n, such that
1. the product 1 6 6
2. the sum 5
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Which two numbers have product 6 and sum 5? By trial and error, you will find that
2 and 3. We use these two numbers to substitute 5 by 2 3 . Thus,
5 6 2 3 6. Then we use compounded:
5 6 2 3 6 2 3 2
2 3
Example
17
12
Since 0 and 0, we must first separate the central term, 17 into two
parts. To do this, we must find two numbers, m and n, such that
1.the product 6 12 72
2. the sum 17
What are the two numbers whose product is 72 and the sum is ‐17 ? You will find that
8 and 9. Note that the negative signs are very important since we want
a sum of ‐17. We use these two numbers to substitute 17 by 8 9 . Thus,
6 17 12 6 8 9 12
6 17 12 6 8 9 12 2 3 4 3 3 4
3 4 2 3
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Example
10
Since 0 and 0, we must first separate the central term, into two parts. To
do this, we need to find two numbers, m and n, such that
1. the product 3 10 30
2. the sum 1
Which two numbers have product ‐30 and sum ‐1? The only possible combination is
6 and 5. Note that the signs must be chosen to correspond to the criteria
of product and sum … We use these two numbers to substitute ‐x by ‐6x + 5x. Therefore,
3 1 12 3 6 5 10
We can now proceed by compounded factorization, i.e. by first finding the common
factors to each pair of terms :
3 is a common factor of 3 6 ;
3 10 3 6 5 10 3 2 5 2
2 3 5
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4. Exercises
1. 4 25
2. 3 12
3. 2 8
4. 3 7 6
5. 3 – 28
6. 12 36
7. 8 7 1
8. 4 4 1
9. 10 3 4
10. 9 20
Solutions
1. 2 5 2 5
2. 3 12 3 4
3. 2 4 2 2 2
4. 3 2 3
5. 7 4
6. 6 6 –6
7. 8 1 1
8. 2 1 2 1 2 –1
9. 2 1 5 4
10. 5 4
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