Difference of two squares.
Lessons on Quadratics: [Summary
- the Program]
Page Overview:
The following column multiplication shows why the difference of two squares
identity hold
(C+A)(C-A) = C2 - A2
Column Multiplication
C + A
C - A (times)
CA + C2
-AC - A2 (add)
C2 - A2
The identity
(C+A)(C-A) = C2 - A2
in equivalent form
C2 - A2 = (C+A)(C-A)
shows how the difference of two squares may be factored.
In practice, the application of this identity to expressions requires
identification of the C2 and - A2 parts of the
expression.
Example (I) - Applied to Numbers
36 - 16
= 62 - 42
C2 - A2
= (6+4)(6-4) (C+A)(C-A)
= (10)(2)
= 20
Example (II) - Applied to Expressions
x2 - 25
= x2 - 52
C2 - A2
= (x+5)(x-5) (C+A)(C-A)
Example (III) - Applied to Expressions
(2x-3)2 - 100
=(2x-3)2 - 102
C2 - A2
= ((2x-3)+10)((2x-3) -10) (C+A)(C-A)
= (2x+7)(2x-13)
The Zero Product Law
Exercise - or food for thought.:
Find the value of the missing digits ABC in the following puzzle
234
ABC (times)
000
000
000 (add)
000
Assumption: If all the factors in a product are nonzero then the
product is nonzero.
The equivalent contrapositive form of this assumed implication rule is the
following: Zero Product Law
If a product equals zero then at least one of its factors equals zero
The zero product law, its direct or indirect assumption, provides a
reason for factoring quadratics and further polynomials p(x) for which the
question where does p(x) = 0 is of interest.
Two ways to solve equations:
Example A.
Suppose we want to solve
0 = x2 - 25 = (x+5)(x-5).
The latter was shown above in example II
The solve an equation route would be to observe
x2 = 25 requires x = 5 or x = -5
since 5 = the principal square root of 25.
The factorization route may be to observe that both factors in the product
(x+5)(x-5) = x2 - 25 will be nonzero when x does not have the
values + 5 or -5. The only way that one of the factors and hence the product can
be zero is if x = 5 or x = -5.
Example B.
Suppose we want to solve
0 = (2x-3)2 - 100 = (2x+7)(2x-13)
The latter was shown above in example III.
The equation route to solve is say
(2x-3)2 = 100 requires
| |
2x -3 = 10 |
or |
2x -3 = -10. |
| Therefore |
|
|
|
| |
2x = 13 |
or |
2x = -7. |
| Therefore |
|
|
|
| |
x = 13/2 = 6.5 |
or |
x =-7/2 = -3.5 |
Check that for x = 6.5 and x = -3.5, both satisfy
0 = (2x-3)2 - 100 or (2x-3)2
= 100
Now the factoring route observes
(2x-3)2 - 100 = (2x+7)(2x-13)
and that the product can only be zero if and only if
2x+ 7 = 0 or 2x - 13 = 0
or equivalently, if and only if,
2x = -7 or 2x = 13
or equivalently, if and only if,
x = -7/2 = -3.5 or x = 13/2 = 6.5
This is the same result as before, except for the order of the solutions, an
immaterial matter.
In solving quadratic and other polynomial equations, you have a choice
between the factorization route and the equation solving route. Choose the route
that is most convenient if that is permitted by your instructor.
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