6.4 Factoring and Solving Polynomial Equations

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6.4 Factoring and Solving Polynomial Equations
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Factor Polynomial Expressions

• In the previous lesson, you factored various
  polynomial expressions.
• Such as
• x3 2x2
• x4 x3 3x2 3x

Grouping common factor the first two terms and
then the last two terms.
Common Factor
x2(x 2)
x(x3 x2 3x 3)
xx2(x 1) 3(x 1)
Common Factor
x(x2 3)(x 1)
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Solving Polynomial Equations

• The expressions on the previous slide are now
  equations
• y x3 2x2 and y x4 x3 3x2 3x
• To solve these equations, we will be solving for
  x when y 0.

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Solve
Let y 0

• y x3 2x2
• 0 x3 2x2
• 0 x2(x 2)
• x2 0 or x 2 0
• x 0 x 2
• Therefore, the roots are 0 and 2.

Common factor
Separate the factors and set them equal to zero.
Solve for x
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Solve
Let y 0

• y x4 x3 3x2 3x
• 0 x4 x3 3x2 3x
• 0 x(x3 x2 3x 3)
• 0 xx2(x 1) 3(x 1)
• 0 x(x 1)(x2 3)
• x 0 or x 1 0 or x2 3 0
• x 0 x 1 x
• Therefore, the roots are 0, 1 and 1.73

Common factor
Group
Separate the factors and set them equal to zero.
Solve for x
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The Quadratic Formula

• For equations in quadratic form ax2 bx c
  0, we can use the quadratic formula to solve for
  the roots of the equation.
• This equation is normally used when factoring is
  not an option.

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Using the Quadratic Formula

• Solve the following cubic equation
• y x3 5x2 9x
• 0 x(x2 5x 9)
• x 0 x2 5x 9 0
• We can, however, use the quadratic formula.

Can this equation be factored?
We still need to solve for x here. Can this
equation be factored?
YES it can common factor.
No. There are no two integers that will multiply
to -9 and add to 5.
a 1 b 5 c -9
Therefore, the roots are 0, 6.41 and -1.41.
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Factoring Sum or Difference of Cubes
If you have a sum or difference of cubes such as
a3 b3 or a3 b3, you can factor by using the
following patterns.
Note The first and last term are cubed and
these are binomials.
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Example
Factor x3 343.
Note This is a binomial. Are the first and
last terms cubed?
x3 343 (x)3 (7)3
( )( - )
x
7
x2
7x
49
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Example
Factor 64a4 27a
a(64a3 27)
Note Binomial. Is the first and last terms
cubes?
a( (4a)3 (3)3)
Note
a( - )( )
4a
3
16a2
12a
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Factor by Grouping
Some four term polynomials can be factor by
grouping.
Example. Factor 3x3 7x2 12x 28
Step 1 Pair the terms.
Step 2 Factor out common factor from each pair.
Identical factors
Step 3 Factor out common factor from each term.
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Example
Factor 3x3 7x2 -12x - 28
Step 1
Note Subtraction is the same as adding a negative
Step 2
Step 3
Note This factor can be further factored
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Solving Polynomial Equations
Solve
Set equation equal to zero.
Factor.
Set each factor equal to zero and solve.