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Math-3
Lesson 2-6
Factoring 3rd Degree Polynomials,
and
Quadratic Form
Quiz 2-51. Find the zeroes by completing the square
2. Find the equation
of the graph (passes
through (0, 7) and (1, 12).
37122 xxy
“Nice” (factorable) 3rd Degree Polynomials
dcxbxaxy 23
)( 2 cbxaxxy
If it has no constant term, it will look like this:
cxbxaxy 23
This can easily be factored (by taking out the common factor ‘x’).
Resulting in ‘x’ times a quadratic factor.
We have been factoring quadratics for quite a while!
Find the zeroes of the following 3rd degree Polynomial
xxxy 45 23
)45(0 2 xxx
)4)(1(0 xxx
Set y = 0
0, -1, -4
xxx 450 23 Factor out the common factor.
Factor the quadratic
Identify the zeroes
“Nice” 3rd Degree Polynomial (with no constant term)
xxxy 45 23
If the quadratic factor is “nice” we can factor that into 2 binomials.
)45( 2 xxxy
It has no constant term so it can easily be factored into ‘x’ times a quadratic factor.
)4)(1( xxxy
This is now “intercept form” so we can “read off” the x-intercepts. What are they?
0, -1, -4
“Nice” 3rd Degree Polynomial (with no constant term)
046 23 xxxy
What if the quadratic factor is not factorable?
)46(0 2 xxx
It has no constant term so it can easily be factored into ‘x’ times a quadratic factor.
Solve by quadratic formula or completing the square!
x = 0
xxx 460 23
49960 2 xx
5)3(0 2 x
53x
Your turn:Factor the following “nice” 3rd degree
polynomials then find the “zeroes” of the polynomial.
xxxy 145 23 xxxy 703 23
)145(0 2 xxx
)2)(7(0 xxx
0, -7, 2
xxx 1450 23
)703(0 2 xxx
)10)(7(0 xxx
0, -7, 10
xxx 7030 23
Another “Nice” 3rd Degree Polynomial
4221 23 xxxy
dcxbxaxy 23
This has the constant term, but it has a very useful feature:
What pattern do you see?
12
13
strd
12
24
24
ndth
6432 23 xxxy
Your turn:
Which of the following 3rd degree polynomials have the “nice pattern” we saw on the previous slide?
151254 23 xxxy
10584 23 xxxy
211876 23 xxxy
22
41
3 nd
th
st
rd
32
41
3 nd
th
st
rd
45
24
13 nd
th
st
rd
32
41
3 nd
th
st
rd
All of them!
4221 23 xxxy
“factor by grouping” if it has this nice pattern.
)42()2( 23 xxxy
Group the 1st and last pair of terms with parentheses.
Factor out the common term from the first group.
)42()2(2 xxxy
Factor out the common term from the last group.
)2(2)2(2 xxxy
Common factor of (x + 2)
2 and 2x
These two binomials are now common factors of
Factor out the common binomial term.
)2)(2( 2 xxy
Apply the Complex Conjugates Theorem
)2(2)2(2 xxxy
2i- ,2i ,2x
Your turn: Find the zeroes using Factor by “grouping”.
151254 23 xxxy
)1512()54(0 23 xxx
)54(3)54(0 2 xxx
)54)(3(0 2 xx
45 ,3i- ,3i x
?42 x )2)(2( ixix
Applying the Complex Conjugates Theorem
))((12 ixixx
22x
?32 x
)2)(2( ixix
)3)(3( ixix
)4)(4( ixix
))((2 aixaixax
What have we learned so far?
cxbxaxy 23 )( 2 cbxaxx
“Nice” Common Factor 3rd degree polynomial:
xxxy 23 23 )23( 2 xxx
)2)(1( xxx
“Nice” Factor by Grouping 3rd degree polynomial:
6432 23 xxxy )32(2)32(2 xxx
)2)(32( 2 xx
)2)(2)(32( ixixx
?814 x
Always factors into: ))(( 22 axax ax 4
)9)(9( 22 xx
“Nice” Difference of Squares:
)3)(3)(3)(3( ixixxx
Looks like23 24 xx
“Quadratic Form”: a trinomial that looks like a
quadratic but has a larger degree.
BUT, according to the Linear Factorization Theorem,
it factors into 4 linear factors.
)1)(2( xx
23 12 xxFactors similarly
))()(2)(2( ixixixix
)1)(2( 22 xx
Your turn
?2816 24 xx )2)(14( 22 xx
)2)(2)(14)(14( xxxx
?8116 4 x )94)(94( 22 xx
)32)(32)(32)(32( ixixxx
)3( x )93( 2 xx
Multiply this
)93(3)93( 22 xxxxx
279393 223 xxxxx
273 x2xThere are NO terms and NO ‘x’ terms
Factoring:
Sum and Difference of 2 Cubes
33 yx
What is the difference of 2 cubes?
Notice: there are
no terms,
no terms,
no terms.
yx2
22 yx2xy
273 x
There are no or x terms.
___)( x )____ ___x ( 2 x
Left times left is left
)3( x )9 ___x ( 2 x
right times
right is right
)3( x x(_____) + = 023xx3
)3( x )93( 2 xx
Now the middle term
of the quadratic
2x
)9 ___x ( 2 x
3(____) + 9x = 0x3
33 yx
__)( x )__ __x ( 2 x
Left times
left is left
)( yx ) __x ( 22 yx
right times
right is right
)( yx x(____) + = 0yx2xy
)( yx )( 22 yxyx
middle term, no
term remains after
multiplying.
) __x ( 22 yx Now check it (no
term remains).
y(____) + = 02xyxy
_)(_ )_ _ (_
yx2
2xy
Factoring: Sum of 2 Cubes
33 yx )( yx )( 22 yxyx
273 x )3( x )93( 2 xx
Both the sum and difference of 2 cubes has a
negative sign in one of the factors.
Your Turn: Factor the expression.
1253 x
648 3 x
)( yx )( 22 yxyx
)5( x )255( 2 xx
)2(8 x )42( 2 xx
)8(8 3 x
Factoring: Difference of 2 Cubes33 yx
)( yx )( 22 yxyx
83 x
)2( x )42( 2 xx
Your Turn: Factor the expression.
273 x
64125 3 x
)( yx )( 22 yxyx
)3( x )93( 2 xx
)45( x )162025( 2 xx
333 45 x
Your Turn: find the “zeroes” of the polynomial.
1253 xy )( yx )( 22 yxyx
)5( 33 xy
)255( 2 xx)5( xya
acbbx
2
42
5x
)1(2
)25)(1(4)5()5( 2 x
2
100255
2
755
2
3255 i
2
355 i
2
355 ,
2
355 ,5
iix
Your Turn: Solve the polynomial.278 3 xy )( yx )( 22 yxyx
)32( 333 xy
)964( 2 xx)32( xya
acbbx
2
42
23x
)4(2
)9)(4(4)6()6( 2 x
8
144366
8
1086
8
3366 i
8
366 i
4
33
4
3 ,
23 i
-x
8
36
8
6 i
Additional Examples
Your turn: factor the following binomials
62 xy
92 xy
)6)(6( xx
)9)(9( xx
)3)(3( xx
If the original problem said to “solve” (find the x-intercepts).
)32)(2( 2 xxy
6432 23 xxxy Previously factored into:
)32)(2( 2 xxy
032 x
32 x
23x
What values of ‘x’ make y = 0?
Subtract 3 (left/right)
Divide by 2 (left/right)
Graph crosses x-axis at x = -3/2
If the original problem said to “solve” (find the x-intercepts).
)32)(2( 2 xxy
6432 23 xxxy Previously factored into:
)32)(2( 2 xxy
022 x
22 x
What values of ‘x’ make y = 0?
Square root (left/right)
Subtract 2 (left/right)
Graph crosses x-axis at imaginary locations.
22 x
2ix
)32)(2( 2 xxy
6432 23 xxxy factors into:
Which has solutions:
23 ,2 ,2 iix
Can we come up with a general way to factor expression like: )2( 2 x
)2)(2( ixix
There’s an term, no “x” term, and then a +2. 2x
Factors as conjugate pairs.
0)2)(2( ixix
In the last problem we got an interesting quadratic form that
factored into imaginary conjugate pairs.
What about the middle ones?
?42 x )2)(2( ixix
Can we use this as a pattern?))((12 ixixx
22x
?32 x
)2)(2( ixix
)3)(3( ixix
)4)(4( ixix
))((2 aixaixax
)32)(2( 2 xxy
General form for the sum of two squares