1 Why
solve by factoring?
1.1 The most fundamental tools
for solving equations are addition, subtraction,
multiplication, and division. These methods work well for equations
like x + 2 = 10 - 2x and 2(x - 4) = 0.
1.2 But what about equations
where the variable carries an exponent,
like x2 + 3x = 8x - 6? This is where factoring comes in. We will use
this equation in the first example.
1.2.1 The
Solve by Factoring process will require four major steps:
1.2.2 1.
Move all terms to one side of the equation, usually the left,
using addition or subtraction.
1.2.3 2.
Factor the equation completely.
1.2.4 3.
Set each factor equal to zero, and solve.
1.2.5 4.
List each solution from Step 3 as a solution to the original equation.
1.2.6 First Example
1.2.7 x2 + 3x = 8x - 6
1.2.9 The
first step is to move all terms to the left using addition and
subtraction. First, we will subtract 8x from each side.
1.2.10 x2
+ 3x - 8x = 8x - 8x - 6
1.2.11 x2
- 5x = -6
1.2.12 Now,
we will add 6 to each side.
1.2.13 x2
- 5x + 6 = -6 + 6
1.2.14 x2
- 5x + 6 = 0.
1.2.15 With
all terms on the left side, we proceed to Step 2.
1.2.17 We
identify the left as a trinomial, and factor it accordingly:
1.2.18 (x
- 2)(x - 3) = 0
1.2.19 We
now have two factors, (x - 2) and (x - 3).
1.2.21 We
now set each factor equal to zero. The result is two subproblems:
1.2.25 Solving
the first subproblem, x - 2 = 0, gives x = 2. Solving the
second subproblem, x - 3 = 0, gives x = 3.
1.2.27 The
final step is to combine the two previous solutions, x = 2 and
x = 3, into one solution for the original problem.
1.2.28 x2
+ 3x = 8x - 6
1.2.30 Examine
the equation below:
1.2.32 If
you let a = 3, then logivally b must equal 0. Similarly, if you
let b = 10, then a must equal 0.
1.2.33 Now
try letting a be some other non-zero number. You should observe
that as long as a does not equal 0, b must be equal to zero.
1.2.34 To
state the observation more generally, "If ab = 0, then
either a = 0 or b = 0." This is an important property of
zero which we exploit when solving by factoring.
1.2.35 When
the example was factored into (x - 2)(x - 3) = 0, this property
was applied to determine that either (x - 2) must equal zero, or (x
- 3) must equal zero. Therefore, we were able to create two equations
and determine two solutions from this observation.
1.2.36 A Second Example
1.2.39 Move
all terms to the left side of the equation. We do this by subtracting
45x from each side.
1.2.40 5x3
- 45x = 45x - 45x
1.2.41 5x3
- 45x = 0.
1.2.43 The
next step is to factor the left side completely. We first note
that the two terms on the left have a greatest common factor of 5x.
1.2.44 5x(x2
- 9) = 0
1.2.45 Now,
(x2 - 9) can be factored as a difference between two squares.
1.2.46 5x(x
+ 3)(x - 3) = 0
1.2.47 We
are left with three factors: 5x, (x + 3), and (x - 3). As explained
in the "Why does it work?" section, at least one
of the three factors must be equal to zero.
1.2.49 Create
three subproblems by setting each factor equal to zero.
1.2.51 2.
x + 3 = 0
1.2.52 3.
x - 3 = 0
1.2.53 Solving
the first equation gives x = 0. Solving the second equation
gives x = -3. And solving the third equation gives x= 3.
1.2.55 The
final solution is formed from the solutions to the three subproblems.
1.2.56 x
= -3, 0, 3
1.2.57 Third Example
1.2.58 3x4 - 288x2 - 1200 =
0
1.2.59 Steps
1 and 2
1.2.60 All
three terms are already on the left side of the equation, so
we may begin factoring. First, we factor out a greatest common factor
of 3.
1.2.61 3(x4
- 96x2 - 400) = 0
1.2.62 Next,
we factor a trinomial.
1.2.63 3(x2
+ 4)(x2 - 100) = 0
1.2.64 Finally,
we factor the binomial (x2 - 100) as a difference between
two squares.
1.2.65 3(x2
+ 4)(x + 10)(x - 10) = 0
1.2.67 We
proceed by setting each of the four factors equal to zero, resulting
in four new equations.
1.2.69 2.
x2 + 4 = 0
1.2.70 3.
x + 10 = 0
1.2.71 4.
x - 10 = 0
1.2.72 The
first equation is invalid, and does not yield a solution. The
second equation cannot be solved using basic methods. (x2 + 4 = 0
actually has two imaginary number solutions, but we will save Imaginary
Numbers for another lesson!) Equation 3 has a solution of x = -10,
and Equation 4 has a solution of x = 10.
1.2.74 We
now include all the solutions we found in a single solution to
the original problem:
1.2.76 This
may be abbreviated as
1.2.79 x2
+ -12x + 27 = 0
1.2.81 x2
+ -12x + 27 = 0
1.2.82 Reorder
the terms:
1.2.83 27
+ -12x + x2 = 0
1.2.85 27
+ -12x + x2 = 0
1.2.86 Solving
for variable 'x'.
1.2.87 Factor
a trinomial.
1.2.88 (3
+ -1x)(9 + -1x) = 0
1.2.89 Subproblem
1
1.2.90 Set
the factor '(3 + -1x)' equal to zero and attempt to solve:
1.2.95 Move
all terms containing x to the left, all other terms to the right.
1.2.96 Add
'-3' to each side of the equation.
1.2.97 3
+ -3 + -1x = 0 + -3
1.2.98 Combine
like terms: 3 + -3 = 0
1.2.99 0
+ -1x = 0 + -3
1.2.101 Combine
like terms: 0 + -3 = -3
1.2.103 Divide
each side by '-1'.
1.2.108 Set
the factor '(9 + -1x)' equal to zero and attempt to solve:
1.2.113 Move
all terms containing x to the left, all other terms to the right.
1.2.114 Add
'-9' to each side of the equation.
1.2.115 9
+ -9 + -1x = 0 + -9
1.2.116 Combine
like terms: 9 + -9 = 0
1.2.117 0
+ -1x = 0 + -9
1.2.119 Combine
like terms: 0 + -9 = -9
1.2.121 Divide
each side by '-1'.
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On 8 Dec 2005, 06:27.