Download MTH55_Lec-23_sec_5-4_Factor_TriNomials

Chabot Mathematics
§5.4 Factor
TriNomials
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
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[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Review § 5.3
MTH 55
 Any QUESTIONS About
• §5.3 → Factoring by GCF and/or
Grouping
 Any QUESTIONS About HomeWork
• §5.3 → HW-18
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Factor (+1)·x2 + bx + c
 Recall the FOIL method of multiplying
two binomials:
F
O
I
L
(x + 2)(x + 5) = x2 + 5x + 2x + 10
= x2 +
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7x + 10
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Factor (+1)·x2 + bx + c
 To factor x2 + 7x + 10, think of FOIL:
The first term, x2, is the product of the
First terms of two binomial factors, so
the first term in each binomial must be x.
 The challenge is to find two numbers p
and q such that p•q = c, and p+q = b
x2 + 7x + 10 = (x + p)(x + q)
= x2 + qx + px + pq
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Factor (+1)·x2 + bx + c
 Need to find two numbers p & q such that
x2 + 7x + 10 = (x + p)(x + q) = x2 + qx + px + pq
 Thus the numbers p and q must be
selected so that their
• PRODUCT is 10
• SUM is 7
 The Factor Pairs for 10 [and their sums]
• 1·10 [11]; (−1)·(−10) [−11]; 2·5 [7];
(−2)·(−5) [−7];
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Bruce Mayer, PE
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Factor (+1)·x2 + bx + c
 In this case, Examination of the “c” term
factor-pairs revealed the desired numbers
p=2 & q=5
 Thus the factorization for x2 + 7x + 10
(x + 2)(x + 5) or (x + 5)(x + 2).
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Bruce Mayer, PE
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Example  FOIL Factoring
 Multiplying binomials uses the FOIL method, and
factoring uses the FOIL method backwards
Product of x and x
x2 .
is
F
L Product of 5 and –7
is
–35.
Sum of the product of outer and inner terms
O
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I
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Factor x2 + bx + c for Positive c
 When the constant term of a
trinomial is positive, look for two
numbers with the same sign. The
sign is that of the middle term:
x2 – 7x + 10 = (x – 2)(x – 5);
x2 + 7x + 10 = (x + 2)(x + 5);
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Bruce Mayer, PE
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Example  Factor x2 + 7x + 12
 SOLUTION: Think
of FOIL in reverse:
(x +
)(x +
)
 We need a constant
term that has a
product of 12 and a
SUM of 7.
 We list some pairs
of numbers that
multiply to 12
Chabot College Mathematics
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Pairs of
Sums of
Factors of 12 Factors
1, 12
2, 6
3, 4
1, 12
2, 6
3, 4
13
8
7
13
8
7
Bruce Mayer, PE
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Example  Factor x2 + 7x + 12
 Since 3  4 = 12 and 3 + 4 = 7,
the factorization of x2 + 7x + 12 is
(x + 3)(x + 4).
 To check we simply multiply the
two binomials.
 CHECK by FOIL:
(x + 3)(x + 4) = x2 + 4x + 3x + 12
= x2 + 7x + 12 
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Example  Factor y2 – 8y + 15
 SOLUTION: Since the constant term is
positive and the coefficient of the middle
term is negative, we look for the
factorization of 15 in which both factors
are negative. Their SUM must be −8.
Pairs of
Factors of 15
Sums of
Factors
–1, –15
–3, –5
–16
–8
Chabot College Mathematics
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Sum of −8
y2 − 8y + 15 =
(y − 3)(y – 5)
Bruce Mayer, PE
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Factor x2 + bx + c for Negative c
 When the constant term of a trinomial is
negative, look for two numbers whose
product is negative. One must be
positive and the other negative:
x2 – 4x – 21 = (x + 3)(x – 7);
x2 + 4x – 21 = (x – 3)(x + 7).
 Select the two numbers so that the number with the
LARGER absolute value has the SAME SIGN
as b, the coefficient of the middle term
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Example  Factor x2 – 5x – 24
 SOLUTION: The
constant term must
be expressed as the
product of negative &
positive numbers.
 Since the sum of the
two numbers must be
negative, the negative
number must have
the greater absolute
value.
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Pairs of
Factors of 24
Sums of
Factors
1, 24
2, 12
3, 8
4, 6
6, 4
8, 3
23
10
5
2
2
5
x2 − 5x − 24 =
(x + 3)(x – 8)
Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor t2 – 32 + 4t
 SOLUTION:
 Rewrite the trinomial
t2 + 4t − 32.
 We need one positive
and one negative
factor. The sum must
be 4, so the positive
factor must have the
larger absolute value
Chabot College Mathematics
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Pairs of
Factors of 32
Sums of
Factors
1, 32
2, 16
4, 8
31
14
4
t2 + 4t − 32 =
(t + 8)(t − 4)
Bruce Mayer, PE
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Example  Two Variables
 Factor: a2 + ab − 30b2
 SOLUTION
 We need the factors of a2 & 30b2
that when added equal ab.
• Those factors are a, and −5b & 6b.
a2 + ab − 30b2 = (a − 5b)(a + 6b)
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Prime Polynomials
 A polynomial that canNOT be factored
is considered prime.
• Example: x2 − x + 7
 Often factoring requires two or more
steps. Remember, when told to factor,
we should factor completely. This
means the final factorization should
contain only prime polynomials.
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Bruce Mayer, PE
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Example  Factor 2x3−24x2+72x
 SOLUTION
Always look first for a common factor.
In this case factor out 2x:
2x(x2 − 12x + 36)
 Since the constant term is positive and
the coefficient of the middle term is
negative, we look for the factorization of
36 in which both factors are negative.
• Their SUM must be −12.
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Bruce Mayer, PE
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Example  Factor 2x3–24x2+72x
 The factorization of
(x2 – 12x + 36) is
(x – 6)(x – 6)
or (x – 6)2
 The factorization of
2x3 – 24x2 + 72x is
2x(x – 6)2
or 2x(x – 6)(x – 6)
Pairs of
Factors of 36
Sums of
Factors
1, 36
2, 18
3, 12
4, 9
6, 6
37
20
15
13
12
2x3 – 242x – 72x = 2x(x – 6 )(x – 6)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
To Factor (+1)·x2 + bx + c
1. Distribute out Common Factors
2. Find a pair of factors that have c as
their product and b as their sum.
a) If c is positive, its factors will have
the same sign as b.
b) If c is negative, one factor will be positive
and the other will be negative. Select the
factors such that the factor with the larger
absolute value has the same sign as b.
3. CHECK by MULTIPLYING
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Bruce Mayer, PE
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Factoring When: LeadCoeff ≠ 1
 Factoring Trinomials of the
Type ax2 + bx + c
• Factoring
with FOIL
• The
Grouping
Method
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor ax2+bx+c by FOIL
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Bruce Mayer, PE
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Example  Factor 3x2 – 14x – 5
1. First, check for a common factor, or GCF for
all Terms. There is none other than 1 or −1.
2. Find the First terms whose product is 3x2.
The only possibilities are 3x and x:
(3x + )(x +
)
3. Find the Last terms whose product is −5.
Possibilities are (−5)(1) & (5)(−1)

Important!: Since the First terms are not
identical, we must also consider the above
factors in reverse order: (1)(−5), & (−1)(5).
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Bruce Mayer, PE
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Example  Factor 3x2 – 14x – 5
4. Knowing that the First and Last products
will check, inspect the Outer and Inner
products resulting from steps (2) and (3)
Look for the combination in which the sum
of the products is the middle term.
(3x – 5)(x + 1) = 3x2 + 3x – 5x – 5
= 3x2 – 2x – 5
Wrong middle term
(3x – 1)(x + 5) = 3x2 + 15x – x – 5
= 3x2 + 14x – 5
Wrong middle term
close
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Bruce Mayer, PE
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Example  Factor 3x2 – 14x – 5
4. Keep Trying the factors of −5.
(3x + 5)(x – 1) = 3x2 – 3x + 5x – 5
= 3x2 + 2x – 5
Wrong middle term
(3x + 1)(x – 5) = 3x2 – 15x + x – 5
= 3x2 – 14x – 5

CORRECT middle term!
Thus
3x2 – 14x – 5 = (3x + 1)(x – 5)
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LdCoeff ≠ 1 Factorization Notes
 Reversing the signs in the binomials
reverses the sign of the middle term
 Organize your work so that you can
keep track of which possibilities you
have checked.
 Remember to include the largest
common factor - if there is one - in the
final factorization.
 ALWAYS CHECK by multiplying
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Bruce Mayer, PE
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Example  Factor 14x + 5 – 3x2
 SOLUTION:
 It is an important problem-solving
strategy to find a way to make problems
look like problems we already know how
to solve. Rewrite the equation in
descending order.
14x + 5 – 3x2 = – 3x2 + 14x + 5
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Bruce Mayer, PE
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Example  Factor 14x + 5 − 3x2
 Starting with −3x2 + 14x + 5
 Factor out the –1:
−3x2 + 14x + 5 = −1(3x2 − 14x − 5)
= −1(3x + 1)(x − 5)
 The factorization of
14x + 5 − 3x2 is −1(3x + 1)(x − 5).
or (−3x − 1)(x − 5) or (3x + 1)(−x + 5)
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Bruce Mayer, PE
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Example  2Vars: 6x2 − xy − 12y2
 SOLUTION: No common factors exist,
we examine the first term, 6x2. There
are two possibilities:
(2x + )(3x + ) or (6x + )(x + ).
 The last term −12y2, has pairs of factors:
12y, −y 6y, −2y 4y, −3y
and
−12y, y −6y, 2y −4y, 3y
as well as each pairing reversed.
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Bruce Mayer, PE
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Example  2Vars: 6x2 − xy − 12y2
 SOLUTION:
 Some trials such as (2x – 6y)(3x + 2y)
and (6x + 4y)(x – 3y), cannot be correct
because (2x – 6y) and (6x + 4y) contain
a common factor, 2.
 Trial
 Product
(2x + 3y)(3x − 4y)
6x2 − 8xy + 9xy − 12y2
= 6x2 + xy − 12y2
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Bruce Mayer, PE
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Example  2Vars: 6x2 − xy − 12y2
 SOLUTION: the Trial (2x + 3y)(3x − 4y)
incorrect, but only because of the sign
of the middle term.
 To correctly factor, simply change the
signs in the binomials.
 Trial
 Product
(2x − 3y)(3x + 4y)
6x2 + 8xy − 9xy − 12y2
= 6x2 − xy − 12y2
The factorization: (2x − 3y)(3x + 4y)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  18m2 – 19mn – 12n2
 SOLUTION
 There are no common factors.
 Factor the first term, 18m2 and get the
following possibilities: 18mm, 9m2m,
and 6m3m.
 Factor the last term, −12n2, which is
negative. The possibilities are:
(−12n)(n), (−n)(12n), (−2n)(6n),
(6n)(−2n), (−4n)(3n) or (−3n)(4n)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  18m2 – 19mn – 12n2
 Look for combinations of factors such that
the sum of the outside and the inside
products is the middle term, (−19mn).
• (9m + n)(2m − 12n) = 18m2 − 106mn − 12n2
• (9m − 12n)(2m + n) = 18m2 − 15mn − 12n2
• (9m − 3n)(2m + 4n) = 18m2 + 30mn − 12n2
• (9m + 4n)(2m – 3n) = 18m2 − 19mn − 12n2
 Thus ANS →
(9m + 4n)(2m − 3n)
Chabot College Mathematics
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correct middle term
Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring by Substitution
 Some times substituting a single
varaible for a (complicated) expression
reveals an easily factored PolyNomial
 Example  factor p2q2 + 7pq + 6
 SOLUTION
 Rewrite using Product-to-Power
Exponent rule → (pq)2 + 7(pq) + 6
 Now engage a substitution
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Bruce Mayer, PE
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Factoring by Substitution
 factor p2q2 + 7pq + 6 = (pq)2 + 7(pq) +
6
 Now to engage a substitution LET
u = pq
 Replace in the Expression pq with u
u2 + 7u + 6
 The Expression in u is easily
FOIL-factored
u2 + 7u + 6 = (u + 6)(u + 1)
Chabot College Mathematics
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Bruce Mayer, PE
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Factoring by Substitution
 Now BACK Substitute u = pq
(u)2 + 7 (u) + 6 = (u + 6)(u + 1)

(pq)2 + 7(pq) + 6 = (pq + 6)(pq + 1)

p2q2 + 7pq + 6 = (pq + 6)(pq + 1)
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
WhiteBoard Work
 Problems From §5.4 Exercise Set
• 30, 44, 66, 82, 92

Shaded Area
Equals
 1 2
2
2 x  4 x   2 x   x 

 2

A  8x  2x  2 x 4   
2
Chabot College Mathematics
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2
2
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All Done for Today
F.O.I.L.
Factoring
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Bruce Mayer, PE
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring ax2+bx+c by Grouping
1. Factor out the largest common factor,
if one exists.
2. Multiply the leading coefficient a and the
constant c; i.e., form the a•c product
3. Find a pair of factors of a•c whose sum is b.
4. Rewrite the middle term, bx, as a sum or
difference using the factors found in step (3).
5. Factor by grouping.
6. Include any common factor from step (1)
and check by multiplying.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor 4x2 – 5x – 6
 SOLUTION
1. First, we note that there is no
common factor (other than 1 or −1).
2. We multiply the leading coefficient, 4
and the constant, −6: (4)(−6) = −24.
3. We next look for the factorization of
−24 in which the sum of the factors is
the coefficient of the middle term, −5.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor 4x2 – 5x – 6
3.
Pairs of
Sums of
Factors of -24 Factors
1, –24
–23
–1, 24
23
2, –12
–10
–2, 12
10
3, –8
–5
–3, 8
5
4, –6
–2
–4, 6
2
Chabot College Mathematics
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We would normally
stop listing pairs of
factors once we
have found the
one we need
Bruce Mayer, PE
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Example  Factor 4x2 – 5x – 6
4. Next, we express the middle term as a
sum or difference using the factors
found in step (3):
−5x = −8x + 3x.
5. We now factor by grouping as follows:
4x2 − 5x − 6 = [4x2 − 8x] + [3x − 6]
= 4x(x − 2) + 3(x − 2)
= (x − 2)(4x + 3)
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor 4x2 – 5x – 6
6. CHECK by FOIL:
(x − 2)(4x + 3) = 4x2 + 3x − 8x − 6
= 4x2 − 5x − 6 

The factorization of 4x2 − 5x − 6 is
(x − 2)(4x + 3).
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Factor 8x3 + 10x2 – 12x
 SOLUTION
1. Factor out the Greatest Common
Factor (GCF), 2x:
8x3 + 10x2 − 12x = 2x(4x2 + 5x − 6)
2. To factor 4x2 + 5x − 6 by grouping, we
multiply the leading coefficient, 4 and
the constant term (−6): 4(−6) = −24.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Factor 8x3 + 10x2 – 12x
3. next look for pairs of factors of −24
whose sum is 5.
Pairs of Factors of −24 Sums of Factors
3, −8
−3, 8
−5
5
4. We then rewrite the 5x in 4x2 + 5x − 6
using:
5x = −3x + 8x
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Factor 8x3 + 10x2 – 12x
5. Next, factor by grouping:
4x2 + 5x − 6 = [4x2 − 3x] + [8x − 6]
= x(4x − 3) + 2(4x − 3)
= (x + 2)(4x − 3)
6. The factorization of the original
trinomial 8x3 + 10x2 − 12x is
2x(x + 2)(4x − 3)
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Bruce Mayer, PE
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Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
4
5
6
y
5
4
3
2
1
x
0
-3
-2
-1
0
1
2
3
4
5
-1
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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[email protected] • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt