Transcript Factoring Cubic Polynomials

6.5
Factoring
Cubic
Polynomials
1/31/2014
Cube: a geometric figure where all
sides are equal.
10 in
10 in
Volume of a cube: side •side•side
V= 10 •10•10
V = 1000 in3
Perfect Cubes
(Volume of a cube whose sides are
whole numbers)
1000 = 103
125 = 53
729 = 93
64 = 43
512 = 83
27 = 33
343 = 73
8 = 23
216 = 63
1 = 13
The side length is the CUBE ROOT of the perfect cube.
Multiply the following:
𝑥 2 − 4𝑥 + 16
4𝑥 2 − 6𝑥 + 9
𝑥+4
4𝑥 2 − 16𝑥 + 64
𝑥 3 − 4𝑥 2 + 16𝑥
𝑥3
𝑥
+
3
+
2𝑥 + 3
12𝑥 2 − 18𝑥 + 27
8𝑥 3 − 12𝑥 2 + 18𝑥
64
(4)3
8𝑥 3
2𝑥 3
+
+
27
(3)3
Summary:
(𝑥 2 −4𝑥 + 16)(𝑥 + 4) = 𝑥 3 + 64 = 𝑥
3
+ (4)3
(4𝑥 2 −6𝑥 + 9)(2𝑥 + 3) = 8𝑥 3 + 27 = 2𝑥
3
+ (3)3
In Reverse: If you were asked to factor:
𝑥 3 + 64 =
𝑥 3 + (4)3 = (𝑥 2 − 4𝑥 + 16)(𝑥 + 4)
Square the 1st
term base
8𝑥 3 + 27 =
Multiply the
Square the 2nd
first and
term base
second base

a  b  a  ab  b
3
2nd term base
2𝑥 3 + (3)3 = (4𝑥 2 −6𝑥 + 9)(2𝑥 + 3)
Factor of the sum of two cubes:
3
First term base
2
2
a  b
Factor:
𝑥 3 − 64 =
𝑥 3 − 4 3 = (𝑥 2 + 4𝑥 + 16)(𝑥 − 4)
8𝑥 3 − 27 =
2𝑥 3 − 3 3 = (4𝑥 2 + 6𝑥 + 9)(2𝑥 − 3)
Factor of the difference of two cubes:

a  b  a  ab  b
3
3
2
2
a  b
Example 1
Factor the Sum or Difference of Two Cubes
a. Factor x 3 + 216 .
SOLUTION
b. Factor 8p 3 – q 3.

a3  b3  a  b a 2  ab  b2
a. x 3 + 216 = x 3 + 63
= ( x + 6) ( x 2 – 6x + 62)
= ( x + 6) ( x 2 – 6x + 36)

Write as sum of two
cubes.
Example 1
Factor the Sum or Difference of Two Cubes
a3  b3  a  ba2  ab  b2 
b. 8p 3 – q 3 = ( 2p) 3 – q 3
= ( 2p – q )[ ( 2p) 2 + 2pq + q2 ]
= ( 2p – q ) (4p2 + 2pq + q2)
Write as difference
of two cubes.
Factor the polynomial.
Checkpoint



a3  b3  a  b a 2  ab  b2 a 3  b3  a  b  a 2  ab  b 2
ANSWER
1. x 3 + 1
( x + 1)( x 2 – x + 1 )
2. 125x 3 + 8
( 5x + 2 ) ( 25x 2 – 10x + 4 )
3. x 3 – 216
( x – 6 ) ( x 2 + 6x + 36 )

Finding Greatest Common Factor
(GCF)
Find the GCF of the terms in the polynomial:
1. 𝑥 3 +𝑥 2
𝑥2
2. 3𝑥 3 − 18
3
2x
3. 2𝑥 3 + 8𝑥 2 − 12𝑥
Example 2
Factor Polynomials completely
a. Factor x 3 – 5x 2 + 6x.
b. Factor 16x 4 – 2x.
SOLUTION
a. x 3 – 5x 2 + 6x = 𝑥(𝑥 2 − 5𝑥 + 6)
GCF: x
Factor out the GCF: x
Factor using Big X
= x ( x – 3) ( x – 2)
b. 16x 4 – 2x = 2𝑥(8𝑥 3 − 1)
Factor out the GCF: 2x
GCF: 2x
a 3  b3  a  b a 2  ab  b 2 
= 2x ( 2x – 1) ( 4x 2 + 2x + 1 )
Checkpoint
Factor Polynomials
Factor the polynomial.
4. x 3 + 2x 2 – 3x
5. 2x 3 – 10x 2 + 8x
ANSWER
x ( x – 1)( x + 3 )
2x ( x – 4 ) ( x – 1 )
Checkpoint
Factor Polynomials
Factor the polynomial.
ANSWER
6. 3x 4 + 24x
3x ( x + 2 ) ( x 2 – 2x + 4 )
7. 54x 4 – 16x
2x ( 3x – 2 ) ( 9x 2 + 6x + 4 )
Homework:
6.5 p.326 #3-6, 14-19,
28-32
6.5
Factoring
Cubic
Polynomials….
cont’d
2/3/2014
Example 4
Factor by Grouping
Factor the polynomial.
a. x 2 ( x – 1 ) – 9 ( x – 1 )
SOLUTION
a. x 2 ( x – 1) – 9 ( x – 1) = ( x 2 – 9) ( x – 1)
= ( x – 3) ( x + 3) ( x – 1)
Factor our (x-1).
a2 – b2 pattern
Example 4
Factor by Grouping
b. x 3 – 2x 2 – 16x + 32
= ( x 3 – 2x 2 ) + ( –16x + 32 ) Group terms.
= x 2 ( x – 2 ) + ( –16 ) ( x – 2 ) Factor each group
using GCF.
= ( x 2 – 16 ) ( x – 2 )
= ( x – 4) ( x + 4) ( x – 2 )
Factor our (x – 2).
a2 – b2 pattern
Checkpoint
Factor by Grouping
Factor the polynomial by grouping.
8. x 2 ( x + 6 ) – 4 ( x + 6 )
9.
x 3 – 4x 2 – 25x + 100
10. x 3 + 3x 2 + 4x + 12
ANSWERS
( x + 6 )( x – 2 ) ( x + 2 )
( x – 4)( x – 5)( x + 5 )
( x + 3)( x 2 + 4)
Example 5
Solve a Cubic Equation by Factoring
Solve 2x 3 – 14x 2 = – 24x.
SOLUTION 2x 3 – 14x 2 + 24x = 0
2x ( x 2 – 7x + 12 ) = 0
2x ( x – 4) ( x – 3 ) = 0
Rewrite in standard form.
Factor common monomial.
Factor trinomial.
2x = 0 or x – 4 = 0 or x – 3 = 0 Use zero product property.
x = 0, x = 4, x = 3
Solve for x.
Example 6
Solve a Cubic Equation by Factoring
Solve x 3 – 6x 2 + 12 = 2x.
SOLUTION
x 3 – 6x 2 – 2x + 12 = 0
Rewrite in standard form.
( x 3 – 6x 2 ) + ( – 2x + 12 ) = 0
Group terms.
x 2( x – 6 ) + ( – 2 ) ( x – 6 ) = 0
Factor each group.
( x 2 – 2 )( x – 6) = 0
or x – 6 = 0
x2 – 2 = 0
+2 +2
+6 +6
x2 = 2
x=6
Use distributive property.
Use zero product property.
Checkpoint
Solve a Cubic Equation by Factoring
Solve the equation by factoring.
ANSWER
13. x 3 + 3x 2 = 4x
– 4, 0, 1
14. 3x 3 – 30x = 9x
+
– 13, 0
15. x 3 + 2x 2 – 3x = 6
– 2, +
– 3
16. x 3 – 7x 2 = 5x – 35
+
– 5, 7
Homework:
WS 6.5 Prac A