Transcript 1.4 Fractional Expressions

Solving Polynomials
1
2
3
Review of Previous Methods
Factoring Cubic Expressions
U-Substitution
Basic Steps
2
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Set the polynomial equal to zero
Factor
Set each factor equal to zero
Solve each factor
CHECK THE ANSWERS!
 Substitute
the answer into the original equation &
verify you get the same value on both sides of the
equal sign
Greatest Common Factor (GCF)
3
Look for the largest common number and/or
variables in each term
 Divide each term by the common terms
Factor:
18 x 4  12 x 2  6 x 5

3 x (6 x  4  2 x )
2
2
3
3x (2 x  6 x  4)
2
3
2
Rainbow Method (2nd degree
Polynomial)
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Grouping
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Solve as you would the last 3 steps of the “Rainbow”
Method
Only works with 4 terms that GCF does not apply to
Steps:
 Write
polynomial in standard form
 Use GCF on the first 2 terms
 Use GCF on the last 2 terms
 Note:
You MUST have repeated expressions in parenthesis
 Otherwise,
it’s prime or was factored incorrectly
 Rewrite the factors outside the parenthesis inside their own
parenthesis
 Write the repeated expression once in parenthesis
Grouping Example
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Factor
a  3a  4a  12
3
2
 a 3  4a 2  3a  12
 a (a  4)  3(a  4)
2
 (a 2  3)(a  4)
Special Cases
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Factoring Formulas
Example
Difference of two Squares:
9a2 – 16
=(3a)2 – (4)2
= (3a + 4)(3a – 4)
x  y  ( x  y )( x  y )
2
2
Difference of two Cubes:
x3 – y3 = (x - y)(x2 + xy + y2)
8a3 – 27
= (2a)3 – (3)3
= (2a – 3)[(2a)2 + (2a)(3) + (3)2]
=(2a – 3)(4a2 + 6a + 9)
Sum of two Cubes:
x3 + y3 = (x + y)(x2 – xy + y2)
125a3 + 1
= (5a)3 + (1)3
= (5a + 1)[(5a)2 – (5a)(1) + (1)2]
Difference of Squares
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
Must have a “-”
x  y  ( x  y)(x  y)
2
2
9a 2  16
 (3a) 2  (4) 2
 (3a  4)(3a  4)
Difference of Cubes
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Must have a “-”
x 3  y 3  ( x  y )( x 2  xy  y 2 )
8a  27
3
 (2a) 3  (3) 3

 (2a  3) (2a) 2  (2a)(3)  (3) 2
 (2a  3)(4a 2  6a  9)

Sum of Cubes
10

Must have a “+”
x 3  y 3  ( x  y )( x 2  xy  y 2 )
125a  64
3
 (5a) 3  (4) 3

 (5a  4) (5a) 2  (5a)(4)  (4) 2
 (5a  4)(25a 2  20a  16)

U-Substitution
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Typically used when other methods are close to
working, but not quite right
Used in calculus…a lot
Steps:
 Replace
the variable with the letter “u” to a power that
would make another method work
 Factor
 Re-substitute the original value for “u”
 Factor again if possible & solve
U-Substitution Example
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
Factor
x 4  3x 2  4
u  x2
x 4  3x 2  4
4 4
( x  4)( x  1)  0
x  3x  4  0
( x  2)( x  2)( x 2  1)  0
4
2
u  x2
u  3u  4  0
2
(u  4)(u  1)  0
2
2