Transcript CH5

2x  6x2
12 x  x 2
4 x  3x 2
Any combination of the prime factorization.
2  2  3 x  x  x
Find the number that “GAZINTA” all the numbers.
6)
2
3
6 goes into 12, 2 times
and into 18, 3 times.
The only number that
divides into 2 and 3 is
1…the number on the
left side is the GCF.
GCF = 6
4)
2) 6
3
4)
8
4
The 6 and the 8 can
still be divided by 2.
The product of the
numbers on the left
side is the GCF.
GCF = 4 * 2 = 8
4
6
9
There are no common
factors in the remaining
factors of 4, 6, and 9.
The GCF = 4
From the previous examples.
6  2x  6  3
6  2 x  3
8  3a  8  4b
4  4x2  4  6x  4  9

8  3a  4b
Leftovers
4  4x2  6x  9
Leftovers
Leftovers
Distributive Property.
When variables are a GCF, it will always be to the smallest power.
3  4x  3 1
34x  1
4ab  a  4ab 10b
4aba  10b
 3x 2 y 2  x 3  5 xy  2xy3  3y 2

6x 2  x 3  4 x  3
x  y  2a  3b 


Group the terms in half.
Factor each side by GCF’s
2
x
1  
1 
5x  ___
3 ___
5x  ___
___ ___
___
Since the 1st two terms are subtracting, both ( )’s will have minus signs.
In fact, both sets of ( )’s must be the same for this to factor. If not the same, prime.
Factor the same binomials as a GCF.
5x 1__x 2 __3
Factor by grouping
2 x3  8x 2  x  4
2
2x
x  ___
x  ___
4  ___
4 
 1 ___
___ ___
x  42x
__2 __
1
8 x 4  6 x  28 x 3  21
2 x ___
7 ___
3  
3 
4x 3  ___
4x 3  ___
___
___
4x
3

7
x __
 3 2__
6ax  3bx  4ay  2by
6w  10wx  35x  21
3 x ___
2 y ___
b  ___
b 
2 a  ___
___
2 a  ___
2 w ___
7 ___
5x  
3  ___
___
___
5x 
3  ___
2a  b3__x  __
2 y
5x  32__w __7 
F.O.I.L.
F  L  10x 2 O  I  10x 2
When the “c” term is positive it
means that the binomials have
the same signs, and the sign on
the “b” term determines the signs
of the binomials.
F  L  10x 2 O  I  10x 2
F  L  10x 2 O  I  10x 2
F  L  10x 2 O  I  10x 2
When the “c” term is negative it
means that the binomials have
the opposite signs, and the sign
on the “b” term determines the
signs of the largest value in the
binomials.
There is another pattern. Find the product of the F & L terms and O & I terms.
Make up any two binomials, with no GCF, and FOIL them.
This pattern gives us the a, b, c rule for finding our factors.
ax 2  bx  c
I  ac
O
___ ___
___
b
I x  bx
O
x  ___
ax 2  Ox  Ix  c
___ ___  ___  ___ ___  ___ 
Factor by Grouping
Factor the trinomials.
x 2  8 x  12
Number Sense!
x 2  8 x  20
x 2  5x  6
___
2  a  c  1 12

6 ___
2  b  8
___
 6___
___  ___
2  a  c  1  20
 10
2  b  8
___  ___
 10
___
 2  a  c  1 6
 3  ___
2  b  5
___
 3  ___
x 2  6 x  2 x  12
x 2  10 x  2 x  20
3 __
2
__
x  __
x  __
 2 ___
6  ___
6
x ___
x  ___
x  ___
___
2
6 __
x  __
x  __
__
Answer looks like.
 2 ___
10  ___
10 
x ___
x  ___
x  ___
___
x  10
x  __
2
__
__ __
Big
Answer looks like.
Did you just notice that the numbers in the binomial answers
are the same numbers that were our factors?
This will always happen when a = 1!
Answer looks like.
Number Sense Rules.
Odd + Even = Odd
Even + Even = Even
Odd + Odd = Even
2 is a factor every Even.
(Odd)*(Odd) = Odd
Factor the trinomials.
 1  a  c  1  6
___
6  ___
 1  b  5
___
6  ___
even + odd
odd
1
6 x  __
x  __
___
___
12  c  24 even

2 

2  ___
12  b  10
___
even
even + even
2 x 12
x  __
__ 
___
 5  ___
 15  c  75 odd
 5  ___
15  b  20
___
even
odd + odd
5 x  15
x  __
__ 
(odd)(even)=even
(odd)*(odd) = odd
___
 3  ___
 7  a  c  1 21
___
___
3  
7  b  4
even
odd + odd
3 x  __
7 
x  __
___
___
4 
6  a  c  1 24
 6  b  10
___
4  ___
even
even + even
4 x  __
6
x  __
___
 8  c  24
 3  ___
___

3  ___
 8  b  5
odd
odd + even
 3  ___
___
 8  c  24
___
 3  ___
 8  b  11
odd
odd + even
8
3 x  __
x  __
8
3 x  __
x  __
___
12  c  48 even
 4___
12  b  8
 4  ___
___
even
even + even
12
4 x  __
x  __
 12  c  60
___

5  ___
12  b  7
 5  ___
___
5(-12) = -60 not 60!
x  __ x  __ 
Prime
Prime doesn’t happen too often, so make sure you check everything!
___

2 ___
18  c  36 even
 18  b  20
___
2  ___
even
even + even
___
 4  ___
 5  c  20
5  b  1
4
___
___
___
6  c  6

1  ___
___
1  ___
 6  b  7
2 x  18
x  __
__ 
5
4 x  __
x  __
6 
1 x  __
x  __
 1  ___
___
8  c  8
 8  b  7
 1  ___
___
1(-8) = -8 not 8!
 3  c  45 odd
___
15  ___
___
15  ___
 3  b  12
 8  c  24
___
3  ___
___
3 
___
8  b  5
x  __ x  __ 
15 y x  __3 y 
x  __
8 n
m  __3 nm  __
Prime
odd + odd
even
These directions means more than one factoring…Watch for GCF!
GCF
of 5

5 x2  x  6
GCF
of x2 2

 3  c  6
___
2  ___
 3  b  1
___
2  ___

12 x 11
x 2 x  __
__ 
 2 x  6 x  18
2

?  ___
?  c  18 even
___
?  ___
?  b  6
___
even
even + even
 2x  __ x  __ 


___
 11  c  132
 12 ___
11  b  1
___
___

12 
3
2 x  __
5x  __
GCF
of -2

x x 2  x  132
GCF
of -3x6
 2 x 2  6 x  18

Not Prime…factored -2 out!
GCF
of 4

 3x 6 x 2  4 x  5

 5  c  5
___
 1 ___
 5  b  4
 1  ___
___
5
1 x  __
 3x 6 x  __

4 x3  2 x 2 y 2  12 y 3

Can’t go any further because
of the variables cubed.
F.O.I.L.
F  L  60x 2 O  I  60x 2
F  L  60x 2 O  I  60x 2
F  L  60x 2 O  I  60x 2
F  L  60x 2 O  I  60x 2
When the “c” term is
positive it means that
the binomials have the
same signs, and the
sign on the “b” term
determines the signs
of the binomials.
When the “c” term is
negative it means
that the binomials
have the opposite
signs, and the sign on
the “b” term
determines the signs
of the largest value in
the FACTORS not
the binomials.
The author actually suggested guessing what the binomials are and FOILing
them out to test if the middle term is correct. 8 tries to get the right answer!?!
Refers to the middle term. ODD + EVEN = ODD
2 5 3 4
___
15  ___
 8  a  c  10  12
___
15 ___
8  b  7
odd
even
10 x 2  15 x  8 x  12
 4 ___
5x ___
2x  ___
3  ___
2x  ___
3
___
4 
2x  __
3 __
5x  __
__
Answer looks like.
It should still factor if we
switch the 15x and -8x.
10 x 2  8 x  15 x  12
 3 ___
2x ___
5x  ___
5x  ___
4  ___
4
___
3 
2x __
4 __
5__x  __
Since we have an odd + even, we need odd factors.
Break the 10 and 12 down to odd factors.
Isolate the odd factors and multiply all possible odd combinations.
3 ___
40
5 ___
24
15 ___
8
3  5  2  4
3  40  7
5  3  2  4
5  24  7
5  3  2  4
15  8  7
Not the factors
Not the factors
Right factors
I can see a pattern! When you look at the left side of each
factoring by grouping, I see the two binomials in the answer!
Do you see that? Say YES! What terms are generating
these binomials? Look above each step.
It is the leading term and the two factors! Can we all agree
that we will always factor out at least an x as the GCF? Yep.
Here is a shortcut. Always put the “a” in both binomials.
Put in the
factors.
Take out
GCF’s
8 10x  __
15 
10x  __
2
5
5x  42x  3
Refers to the middle term. EVEN + EVEN = EVEN
222 3 5
Since we have an even + even, we factor out a 2 from our factors.
Break the 8 down to get factors of 2’s. Put a 2( ) in each blank
as a factor because we know that the two factors are even.
 20  a  c  8  15
6  _____
_____
2(
_____
3 ) 2(_____
 10 )  b  14  2 ( – 7 ) Factor 2 out of the -14. The sum of the two red ( )’s must = (– 7).
even
even
Because a = 8
6 and -20 are the two factors
that add up to -14. Place
them in our answer.
Answer looks like this using the
new short cut. Use 8x twice.
6 8x  20
8x  __
__ 
2
4
4x  32x  5
Since – 7 is odd. Isolate the odd factors and multiply all possible odd
combinations.
3 ___
10
3  2  5
3 10  7
5 ___
15 ___
Right factors! Put them in the red ( )’s!
Now we know we are not finished because we used the 8 twice.
We have to divide out the extra 8 by finding the GCF of each
binomial.
Refers to the middle term. ODD + ODD = EVEN
EVEN RULE ( odd + odd )
Factor. 3 x 2  34 x  63
337
 7  _____
 27  a  c  3  63
_____
_____
 7  _____
 27  b  34
odd
odd
Because both a & c are odd
-7 and -27 are the two
factors that add up to -34.
Place them in our answer.
Since a and c as odd factors we have an odd + odd = -34.
This is going to take some time because all the factors will be odd.
Break 63 down.
Isolate each odd factor, from smallest to largest, and then multiply
all possible odd combinations to create more odd factors.
3 ___
7  ___
9  ___
63
27
3  3  3  7 7  3  3  3
3  63  34  7  27  34
Wrong factors!
21 ___
27  ___
Right factors!
Answer looks like this using
new short cut. Use 3x twice.
7 3x  27
3x  __
__ 
NO
GCF
3
3x  7x  9
Now we know we are not finished because we used the 3 twice.
We have to divide out the extra 3 by finding the GCF of each
binomial.
These directions means more than one factoring…Watch for GCF!
2 3
2 4 The 2 or 4 must
_____
 2  _____
 12  a  c  3  8
 2  _____
 12  b  10
_____
be multiplied to
the 3
even
even
Because c = -8
odd
2 3x 12
3x  __
__ 
NO
GCF
GCF
of 2x.

3
3x  2x  4
Remember this example 2 pages
ago, where the author FOILed it
out 8 times? Which way is easier?

2 x 4 x  11x  3
2
Because 11 is
much bigger
_____
 1  _____
 12  a  c  4  3 than 4 and 3,
multiply 4 and 3
 1  _____
 12  b  11
_____
to get 12.
odd
even
2x4x 12
__ 4x  __
1
4
NO
GCF
_____
 3  _____
 10  a  c  5  6
 3  _____
 10  b  7
_____
2 xx  3 4x 1
The 5 and -6
doesn’t work, so
try 3 and 10!
-3 and 10 work.
even
3
5x 10
__ 5x  __
x  25x  3
NO
GCF
5
22
One of the 2’s must
_____
 10  a  c  4  5 be multiplied to the
 2  _____
 10  b  12 5. 2 and 10
 2  _____
_____
even
even
Because a = 4
2 4x  __
10
4x  __
2
2
2 x  12x  5
Factor completely.
Need to have x powers in
descending order.
6 x 2  4 y 2  19 xy
6 x 2  19 xy  4 y 2
63 is a big value…
factors must be far apart.
10 x 2  63x  18
2 3
_____
 3  _____
 60  a  c  10 18
 3  _____
 60  b  63
_____
1
 24
_____ _____  a  c  6  4
3
8
_____  _____  b  19 No possible factors.
odd
even
odd
PRIME
6x  __ y 6x  __ y
6 x  23x  20
2
_____
 a  c  6  20
15  _____
8
_____
 b  23
15  _____
8
odd
even
8 
6x  15
__ 6x  __
3
2
odd
even
odd
40
3*_____
24
5*_____
15*____
8
2x  5 3x  4
odd
60
3*_____
5*_____
9* ____
15*____
45*____
3 10x  __
60 10 x  3x  6
10x  __
NO
GCF
2 3 225
2 5 233
10
222
One of the 3’s
3 3 must be mult. to
8x  6x  9
 6  _____
12  a  c  8  9 the 2, 6 and 3
_____
subtract to be
_____

_____

b

6

3
6
 23 the 3 in the ( )’s.
2(
) 2(
)
2
even
even
6
8x  12
__ 8x  __
4
2
even
2x  3 4x  3
Factor completely.
GCF 30 x 4  28 x 3  30 x 2
of 2x2.

2 x 15 x  14 x  15
2
2

GCF
2
of -3.  18 x  39 x  18
3 5 3 5

 3 6 x  13x  6
_____  _____  a  c  15  15
75
3*_____
_____  _____  b  14
odd
odd
2 x 2 15 x  __ 15 x  __ 

2 x 15 x  14 x  15
2
2
6 x 2  17 xy  28 y 2
even
45
5*_____
9*____
25

even
7 y
6x  24
__ y 6 x  __
6
NO
GCF
x  4 y  6 x  7 y 
odd
odd
even
24
7*_____
21*____
8
odd
9 6 x  __
4
 36 x  __
2
 32 x  3 3x  2
2 3 227

3 2 3 2
_____
 9  _____
 4  ac  66
 4  b  13
_____
 9  _____
3
_____
 7  _____
 24  a  c  6  28
56
3*_____
 24  b  17
_____
 7  _____
odd
2
223
12 x 2  28 x  15
The two 3’s
mult. together, 9
and 5 add up to
be the -14 in the
( )’s.
3 5
_____
 18  _____
 10  a  c  12 15
2( 9)  _____
2(  5)  b  28 214
_____
even
even
even
12x  18
__ 12x 10
__ 
6
2
2x  3 6x  5
It is important to know that x2, x4, xeven, etc. are all perfect squares
Per.
Sqr.
x2  4x  4
x  2
2

Per.
Sqr.
Let’s try a ( )2
We must test the middle term!
 2x  2x  4x It Factors!
x  52
x  4 2


Done!
5x  5x  10x
x x__13x2 __
9
Done!
 4x  4x  8x
Done! NOT

Done!
3x  3x  6x
ERASE
4 and 36 are perfect squares,
but 4 is a GCF!

4 x  6x  9
2
4x  3

2

Done!
 3x  3x  6x
4 x  3
3x  7 


2
Done!
12x 12x  24x
2
Done!
 21x  21x  42x
HIDE inside other polynomials!
x  5x  5
x  8x  8
GCF of 9, 1st!


9 x2  4
9x  2x  2
Factor completely.
16 x 4  81
4x
4x
2
2

 9 4x2  9
x2  4

12 x 2  75
NOT Diff. of Per. SQ!
PRIME
Another Diff. of Per. SQ!

 9 2 x  32 x  3
GCF of 3, 1st!


3 4 x 2  25
32 x  52 x  5
Or put the 25y2 in front.
 36 x 2  25 y 2
GCF of -1, 1st!

 1 36 x  25 y
2
2
25 y 2  36 x 2

16 x  5 y 6 x  5 y 
5 y  6x5 y  6x
100a 4c 6  121x8 y 2
Even powers on the variables are
still perfect squares. Divide the
powers by 2 to take the square root.
10a c
2 3

 11x 4 y 10a 2c3  11x 4 y

Middle terms exist!
Binomial
3
Trinomial
Same sign Opposite Always Plus
as given sign as given


b ___
a ___
ab  ___
a 2 ___
b2
a  b  ___

3 3
3



3
3


3 y  __
2x____
___
4x 2 
9 y 2  ____
6 xy  ____



2
20 x  ____
25 
4x  ___
5 16x
___
____  ____

3
3
3
2
 4 2
2
36 
x
y
6
x
y  ____
6
x y  __ ____  ____
___


2a 3 8x3 3 t 3
3
3

3


2
70 x  ____
49 
x  ___
7 100x
10
___
____  ____

1st
Both Diff. of Per. Sq.
and Perfect Cubes
2nd
3
8x


 1 8x3  1
33


t ___
2a___

__
4x 2 2___
xt  ___
t2 
2x

 
1 ___
2 x  __
1 ___
1
x  __
4x 2  ___
2__x  __
1  2__
4x 2  ___
2 x  __

GCF( LEFTOVERS )
The number of terms in the leftovers determines which step we go to.
a 2  b 2  a  ba  b
1st
Remember these like to hide inside of other polynomials.
2nd
3
Same sign Opposite Always Plus
as given sign as given


b ___
a ___
ab  ___
a 2  ___
b2
a  b  ___

3 3
3



Remember the sign rules for what your answer looks like.
e  ac
d  _____
_____
d  _____
e b
_____
d ax  __
ax  __
e
GCF
GCF
3 to 1 SPLIT
a 2 x 2  2abx  b 2  y 2
ax 3  bx 2  cx  d
___ ___  ___  ___
RT ___  ___ 
GCF
LT
GCF
SAME
SAME
___  ___ ___  ___ 
GCF
LT
SAME
GCF
RT
ax  b 2  y 2
Difference of
Perfect Squares
1 to 3 SPLIT
is the same
concept, but
watch for
signs!
GCF
of 5
Step 2
D.P.S.
Step 2
D.P.S.
Again

5 x  16

4

GCF
of 2x

5 x 4 x 4
2
2



2 xx 2 x  5  92 x  5
2 x2 x  5x  9
2 x 2 x 3  5x 2  18x  45
Step 4
F. by G.
2
2


5 x 2  4 x  2x  2
Step 2
D.P.S.
2 x2 x  5x  3x  3
Step 2
D.P.S.


x  8x  8
Step 2
Step 4
& P.C.
F. by G. 3x x  1  8x  1
twice x  2x  2 x  4x  2x
3x  1x  8
Step 2
D.P.S.
3x  1x  1x  2x  2 x  4
& P.C.
GCF
of 3
3 x5  x3  8x 2  8
3
2
3
2
2
2
3
3
2
2
 2x  4

Factor completely.
GCF
of 7
Step 3

7 x  5x  6
2

GCF
of 3x2
7x  3x  2
Step 3
3 to 1 SPLIT
Step 4
F. by G.
 x  3
2
y
Difference of
Per. Squares

3x 2 x 2  10 x  24

3x 2 x  12x  2
1 to 3 SPLIT
Step 4
F. by G.
2
x  3  y x  3  y 
x  3  yx  3  y 

y  x  6x  9
2
y 2  x  3
2
2

GCF of -1
first.
Difference of
Per. Squares
 y  x  3 y  x  3
Distribute the minus!
 y  x  3 y  x  3
Remember the product of -8(10)(3)(5)(0)(7)(11) = 0.
5x  0
x2  0
 2  2
Solve 5
5
each
x0 x2
for x.
x  0,2
x  4x  6  0
x4  0
x4
7
 0 x  8  0 2x  3  0
x6  0
x  4,6
x6
x  8
3
x  8,
2
2x  3
3
x
2
2x  0 x 1  0 x  1  0
x0
x  1 x  1
x  0,1,1
x 12x  2  0
x  12 x  2
x  12,2
Solve the equations by factoring.
x 2  8 x  16  0
x  4x  4  0
x4
We don’t have to list the same
number twice, but just know that
there were two answers that were
the same value.
Never divide by the variable!
Set the equation = 0.
x2  6x
4 x 2  25
x2  6x  0
xx  6  0
4 x 2  25  0
x0 x6
x  0,6
2x  52x  5  0
2x  5  0
2x  5
5
x
2
5
x
2
5 5
x  ,
2 2
No reason to work out the 2nd
binomial because the only
difference will be the sign.
Solve the equations by factoring.
x 2  16 x  63  0
x  7x  9  0
x  7 x  9
x  7 , 9
2


2 x x 2  4 x  12  0
2xx  2x  6  0
x 0 x2
x  6
x  0,2,6
x  14 x  2
x  14,2
2 x  8 x  24 x  0
3
x 2  12 x  28
x 2  12 x  28  0
x  14x  2  0
6 x  x  35  0
2
2 3
75
_____
 15  _____
14  a  c  6  35
_____
 15  _____
14  b  1
odd
even
odd
6x  14
__ 6x 15
__   0
2
3
The factors
have to
differ by 1,
so 2(7)=14
and 3(5)=15
3x  72x  5  0
3x  7  0
3x  7
2x  5  0
2x  5
7
x
3
5
x
2
Solve the equations by factoring.
One of the
must be
8 x 2  30 x  27  0 2 2 2 3 3 33’s
isolated, 3
 6  a  c  8  27 and 18 will
36  _____
_____
subtract to
2(
)
2(
)
_____
18  _____
 3  b  30  2 15 be 15 in the
even
even
even
( )’s.
 
368x  __
6  0
8x  __
4
2
2x  9  0 4x  3  0
2 x  9
9
x
2
4x  3
3
x
4
9 3
x ,
2 4
2 x 3  3x 2  8 x  12  0
x 2 2 x  3  42 x  3  0
2 x  3x 2  4  0
2x  3x  2x  2  0
2x  3  0
x  2
2x  3
3
x
2
3
x  2,2,
2
x2
Solve the equations by factoring.
x  32x 1  9
Will need to FOIL and set = 0
2x2  x  6x  3  9
2 x 2  5 x  12  0
223
_____
 3  _____
8  a  c  2  12
_____
 3  _____
8 b5
odd
even
odd
8 2x  __
3  0
2x  __
x  3x  4  5x
x 2  4 x  3x  12  5 x
x 2  1x  12  5 x
x 2  4 x  12  0
x  6x  2  0
x  6 x  2
2
x40
x  4
2x  3  0
2x  3
3
x
2
3
x  , 4
2
x  6,2
The cutting board is a rectangle because of the reference to “long and wide.” Build a rectangle.
The Area formula is L * W and the Area equals 800.
W
L W  Area
L  2W
2W W  800
We know that the Length
is twice the Width.
2W 2  800
2W 2  800  0


2 W 2  400  0
2W  20W  20  0
W  20 W  20
L  2W
L  2  20  40
The dimensions are 40 cm by 20 cm.
Multiply the two numbers and set = 156
Means that the numbers differ by 1.
The First number is unknown, call it x.
x  x  1  156
The Second number must be 1 bigger… x + 1


x 2  1x  156
Assuming the racing number must be positive,
the first number is 12 and the consecutive
second number is 13.
x 2  1x  156  0
x 13x 12  0
x  13 x  12
Means that the numbers differ by 2.
The First number is unknown call it x.
The Second number must be 2 bigger… x + 2
x  x  2  440
x 2  2 x  440
There are two sets of answers!
x 2  2 x  440  0
x  22x  20  0
x  22 x  20
-22 and -20
20 and 22
c
a
a b  c
The legs of the right triangle are the sides of the right
angle, labeled a and b. The hypotenuse is the longest side
and is labeled c.
b
x 9
Right triangles have a special relationship called The
Pythagorean Theorem.
2
2
2
15
a2  b2  c2
x 2   x  3  152
2
x3
9  3  12
The other two sides
are 9 ft and 12 ft.
x 2  x 2  3x  3x  9  225
2 x 2  6 x  216  0


2 x 2  3x  108  0
2x  12x  9  0
x  12 x  9
N  210
 10t 2  100t  210
 10t  100t  210  0
 10 t 2  10t  21  0
2


10t  3t  7  0
t 3
t 7
There will be 210 micrograms in the
bloodstream at 3 minutes and 7 minutes.
135
352  120 2  c 2
1225  14400  c 2
c
35
0  c  125c 125
15625  c 2
100
120
0  c 2  15625
c  125
The minimum length of the cable is 125 ft.
h 2  h  10  50 2
2
50
h  10
h 2  h 2  10h  10h  100  2500
h
2h 2  20h  2400  0


2 h 2  10h  1200  0
2h  40h  30  0
h  30
The two distances are
30 ft. and 40 ft.
A number is 6 less than its square. Find all such numbers.
x
2
 x 6
0  x2  x  6
0  x  2x  3
x  2
x3
x  x2  6
2
 2   2   6
2  46
x  x2  6
3  3  6
3 96
2