, SOLUTION MANUAL FOR Algebra & Trigonometry
Graphs and Models, A Right Triangle Approach, 7th
edition Bittinger
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
,Just-in-Time Review
√ √ √ √
4. 6 = 6, so it is true that 6 ≤ 6.
1. Real Numbers
5. −30 is to the left of −25 on the number line, so it is false
that −30 > −25.
2 √ 8 4 16 5 25 16 25
1. Rational numbers: , 6, −2.45, 18.4, −11, 3 27, − , 6. − =− and − = − ; − is to the right of − ,
√ 3 7 5 20 4 20 20 20
0, 16 4 5
so it is true that − > − .
2 8 5 4
2. Rational numbers but not integers: , −2.45, 18.4, −
3 7
√ √ √ √
3. Irrational numbers: 3, 6 26, 7.151551555 . . . , − 35, 5 3 4. Absolute Value
(Although there is a pattern in 7.151551555 . . . , there is
no repeating block of digits.)
√ √ 1. | − 98| = 98 (|a| = −a, if a < 0.)
4. Integers: 6, −11, 3 27, 0, 16
2. |0| = 0 (|a| = a, if a ≥ 0.)
√ √
5. Whole numbers: 6, 3 27, 0, 16
3. |4.7| = 4.7 (|a| = a, if a ≥ 0.)
6. Real numbers: All of them
2 2
4. − = (|a| = −a, if a < 0.)
3 3
2. Properties of Real Numbers 5. | − 7 − 13| = | − 20| = 20, or
|13 − (−7)| = |13 + 7| = |20| = 20
1. −24 + 24 = 0 illustrates the additive inverse property. 6. |2 − 14.6| = | − 12.6| = 12.6, or
2. 7(xy) = (7x)y illustrates the associative property of mul- |14.6 − 2| = |12.6| = 12.6
tiplication.
7. | − 39 − (−28)| = | − 39 + 28| = | − 11| = 11, or
3. 9(r − s) = 9r − 9s illustrates a distributive property. | − 28 − (−39)| = | − 28 + 39| = |11| = 11
4. 11 + z = z + 11 illustrates the commutative property of 3 15 6 15 21 21
addition. 8. − − = − − = − = , or
4 8 8 8 8 8
5. −20 · 1 = −20 illustrates the multiplicative identity prop- 15
− − 3 = 15 + 6 = 21 = 21
erty. 8 4 8 8 8 8
6. 5(x + y) = (x + y)5 illustrates the commutative property
of multiplication.
5. Operations Using Fraction Notation
7. q + 0 = q illustrates the additive identity property.
1 1 3 1 4 3 5 4 15 4 + 15 19
8. 75 · = 1 illustrates the multiplicative inverse property. 1. + = · + · = + = =
75 5 4 5 4 4 5 20 20 20 20
9. (x+y)+w = x+(y+w) illustrates the associative property 3 1 3 1 5 3 5 8 4·2 4
2. + = + · = + = = =
of addition. 10 2 10 2 5 10 10 10 5·2 5
10. 8(a + b) = 8a + 8b illustrates a distributive property. 3 5 3 3 5 4 9 20 29
3. + = · + · = + =
8 6 8 3 6 4 24 24 24
5 5 9 5 27 32
4. +3= +3· = + =
3. Order on the Number Line 9 9 9 9 9 9
7 3 7 3 2 7 6 1
5. − = − · = − =
1. 9 is to the right of −9 on the number line, so it is false 8 4 8 4 2 8 8 8
that 9 < −9. 10 2 10 3 2 11 30 22 8
6. − = · − · = − =
2. −10 is to the left of −1 on the number line, so it is true 11 3 11 3 3 11 33 33 33
that −10 ≤ −1. 3 7 3 14 3 11
7. 2 − =2· − = − =
√ √ √ 7 7 7 7 7 7
3. −5 = − 25, and − 26 is to the left of√− 25, or −5, on
the number line. Thus it is true that − 26 < −5. 5 3 5·3 15 5·3 3
8. · = = = =
8 10 8 · 10 80 5 · 16 16
Copyright
c 2025 Pearson Education, Inc.
,2 Just-in-Time Review
3 5 3·5 15 18. First we write each mixed numeral as a fraction.
9. · = =
4 8 4·8 32 1 1 5 1 5 1 6
1 =1+ =1· + = + =
5 2 5 3 15 3·5 5 5 5 5 5 5 5 5
10. ÷ = · = = = 1 1 4 1 8 1 9
6 3 6 2 12 3·4 4 2 =2+ =2· + = + =
4 4 4 4 4 4 4
3 4 48
11. 12 ÷ = 12 · = = 16 Then we divide.
4 3 3 6 9 6 4 24 8·3 8
÷ = · = = =
5 5 4 5 9 45 15 · 3 15
5 3 5 4 20 10 · 2 10
12. 6 = ÷ = · = = =
3 6 4 6 3 18 9·2 9
4 6. Operations with Real Numbers
1 1 6 1 30 1 31
13. 5 =5+ =5· + = + =
6 6 6 6 6 6 6 1. 8 − (−11) = 8 + 11 = 19
14. We divide.
3 1 3·1 3 1 1 1
2. − · − = = · =1· =
3 10 3 10 · 3 3 10 10 10
9 32 32 5
=3 3. 15 ÷ (−3) = −5
−27 9 9
5 4. −4 − (−1) = −4 + 1 = −3
1 5. 7 · (−50) = −350
15. First we write 4as a fraction.
2 6. −0.5 − 5 = −0.5 + (−5) = −5.5
1 1 2 1 8 1 9
4 =4+ =4· + = + = 7. −3 + 27 = 24
2 2 2 2 2 2 2
Then we add. 8. −400 ÷ −40 = 10
1 2 9 2 9 3 2 2 27 4 31 9. 4.2 · (−3) = −12.6
4 + = + = · + · = + =
2 3 2 3 2 3 3 2 6 6 6
10. −13 − (−33) = −13 + 33 = 20
31
Finally we divide to write as a mixed numeral.
6 11. −60 + 45 = −15
5 1 2 1 2 3 4 1
6 31 12. − = + − = + − =−
31 1 2 3 2 3 6 6 6
=5
−30 6 6 13. −24 ÷ 3 = −8
1
14. −6 + (−16) = −22
16. First we write each mixed numeral as a fraction.
1 5 1 8 1·8 1·2/·4 4
5 5 9 5 18 5 23 15. − ÷ − =− · − = = =
2 =2+ =2· + = + = 2 8 2 5 2·5 2·5
/ 5
9 9 9 9 9 9 9
1 1 3 1 3 1 4
1 =1+ =1· + = + = 7. Interval Notation
3 3 3 3 3 3 3
Then we subtract.
1. This is a closed interval, so we use brackets. Interval no-
23 4 23 4 3 23 12 11 tation is [−5, 5].
− = − · = − =
9 3 9 3 3 9 9 9
2. This is a half-open interval. We use a parenthesis on
11
Finally we divide to write as a mixed numeral. the left and a bracket on the right. Interval notation is
9 (−3, −1].
1
9 11 3. This interval is of unlimited extent in the negative direc-
11 2
=1 tion, and the endpoint −2 is included. Interval notation is
−9 9 9 (−∞, −2].
2
4. This interval is of unlimited extent in the positive direc-
1 tion, and the endpoint 3.8 is not included. Interval nota-
17. First we write 3 as a fraction.
3 tion is (3.8, ∞).
1 1 3 1 9 1 10
3 =3+ =3· + = + = 5. {x|7 < x}, or {x|x > 7}.
3 3 3 3 3 3 3
Then we multiply and simplify. This interval is of unlimited extent in the positive direction
1 9 10 9 90 and the endpoint 7 is not included. Interval notation is
3 · = · = =3 (7, ∞).
3 10 3 10 30
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 3
6. The endpoints −2 and 2 are not included in the interval, 2. Convert 0.000786 to scientific notation.
so we use parentheses. Interval notation is (−2, 2). We want the decimal point to be positioned between the
7. The endpoints −4 and 5 are not included in the interval, 7 and the 8, so we move it 4 places to the right. Since
so we use parentheses. Interval notation is (−4, 5). 0.000786 is between 0 and 1, the exponent must be nega-
tive.
8. The interval is of unlimited extent in the positive direc- 0.000786 = 7.86 × 10−4
tion, and the endpoint 1.7 is included. Internal notation
is [1.7, ∞). 3. Convert 0.0000000023 to scientific notation.
9. The endpoint −5 is not included in the interval, so we use a We want the decimal point to be positioned between the
parenthesis before −5. The endpoint −2 is included in the 2 and the 3, so we move it 9 places to the right. Since
interval, so we use a bracket after −2. Interval notation is 0.0000000023 is between 0 and 1, the exponent must be
(−5, −2]. negative.
0.0000000023 = 2.3 × 10−9
10. This interval is of unlimited
√ extent in the negative direc-
tion, and the√endpoint 5 is not included. Interval nota- 4. Convert 8,927,000,000 to scientific notation.
tion is (−∞, 5). We want the decimal point to be positioned between the
8 and the 9, so we move it 9 places to the left. Since
8,927,000,000 is greater than 10, the exponent must be
8. Integers as Exponents positive.
8, 927, 000, 000 = 8.927 × 109
1 1 5. Convert 4.3 × 10−8 to decimal notation.
1. 3−6 = Using a−m =
36 am
The exponent is negative, so the number is between 0 and
1 −m 1 1. We move the decimal point 8 places to the left.
2. = (0.2)5 Using a = m
(0.2)−5 a
4.3 × 10−8 = 0.000000043
w−4 z9 a−m bn
3. −9
= 4 Using −n = m 6. Convert 5.17 × 106 to decimal notation.
z w b a
2 The exponent is positive, so the number is greater than
z z2 10. We move the decimal point 6 places to the right.
4. = 2 Raising a quotient to a power
y y 5.17 × 106 = 5, 170, 000
5. 1000 = 1 Using a0 = 1, a = 0 7. Convert 6.203 × 1011 to decimal notation.
a5 The exponent is positive, so the number is greater than
6. = a5−(−3) = a5+3 = a8 Using the quotient rule
a−3 10. We move the decimal point 11 places to the right.
7. (2xy 3 )(−3x−5 y) = 2(−3)x · x−5 · y 3 · y 6.203 × 1011 = 620, 300, 000, 000
= −6x1+(−5) y 3+1 8. Convert 2.94 × 10−5 to scientific notation.
4
6y
= −6x−4 y 4 , or − The exponent is negative, so the number is between 0 and
x4 1. We move the decimal point 5 places to the left.
1 2.94 × 10−5 = 0.0000294
8. x−4 · x−7 = x−4+(−7) = x−11 , or
x11
1
9. (mn)−6 = m−6 n−6 , or 10. Order of Operations
m6 n6
1
10. (t−5 )4 = t−5·4 = t−20 , or
t20 1. 3 + 18 ÷ 6 − 3 = 3 + 3 − 3 Dividing
= 6−3=3 Adding and subtracting
9. Scientific Notation 2. = 5 · 3 + 8 · 32 + 4(6 − 2)
= 5 · 3 + 8 · 32 + 4 · 4 Working inside parentheses
1. Convert 18,500,000 to scientific notation. = 5·3+8·9+4·4 Evaluating 32
We want the decimal point to be positioned between the = 15 + 72 + 16 Multiplying
1 and the 8, so we move it 7 places to the left. Since = 87 + 16 Adding in order
18,500,000 is greater than 10, the exponent must be posi-
tive. = 103 from left to right
18, 500, 000 = 1.85 × 107
Copyright
c 2025 Pearson Education, Inc.
,4 Just-in-Time Review
3. 5[3 − 8 · 32 + 4 · 6 − 2] 3. 2a4 − 3 + a2
= 5[3 − 8 · 9 + 4 · 6 − 2] The term of highest degree is 2a4 , so the degree of the
= 5[3 − 72 + 24 − 2] polynomial is 4.
= 5[−69 + 24 − 2] 4. −41 = −41x0 , so the degree of the polynomial is 0.
= 5[−45 − 2]
5. 4x − x3 + 0.1x8 − 2x5
= 5[−47]
The term of highest degree is 0.1x8 , so the degree of the
= −235 polynomial is 8.
4. 16 ÷ 4 · 4 ÷ 2 · 256 6. x − 3 has two terms. It is a binomial.
= 4 · 4 ÷ 2 · 256 Multiplying and dividing
7. 14y 5 has one term. It is a monomial.
in order from left to right
= 16 ÷ 2 · 256 1
8. 2y − y 2 + 8 has three terms. It is a trinomial.
= 8 · 256 4
= 2048
5. 26 · 2−3 ÷ 210 ÷ 2−8 12. Add and Subtract Polynomials
−8
= 2 ÷2
3 10
÷2
−7 −8
=2 ÷2 1. (8y − 1) − (3 − y)
=2 = (8y − 1) + (−3 + y)
6. 4(8 − 6)2 − 4 · 3 + 2 · 8 = (8 + 1)y + (−1 − 3)
31 + 190 = 9y − 4
4 · 22 − 4 · 3 + 2 · 8
= Calculating in the 2. (3x2 − 2x − x3 + 2) − (5x2 − 8x − x3 + 4)
3+1
numerator and in = (3x2 − 2x − x3 + 2) + (−5x2 + 8x + x3 − 4)
the denominator
4·4−4·3+2·8 = (3 − 5)x2 + (−2 + 8)x + (−1 + 1)x3 + (2 − 4)
= = −2x2 + 6x − 2
4
16 − 12 + 16
= 3. (2x + 3y + z − 7) + (4x − 2y − z + 8)+
4
4 + 16 (−3x + y − 2z − 4)
= = (2 + 4 − 3)x + (3 − 2 + 1)y + (1 − 1 − 2)z+
4
20 (−7 + 8 − 4)
=
4 = 3x + 2y − 2z − 3
= 5
4. (3ab2 − 4a2 b − 2ab + 6)+
7. 64 ÷ [(−4) ÷ (−2)] = 64 ÷ 2 = 32
(−ab2 − 5a2 b + 8ab + 4)
8. 6[9 − (3 − 2)] + 4(2 − 3) = (3 − 1)ab2 + (−4 − 5)a2 b + (−2 + 8)ab + (6 + 4)
= 6[9 − 1] + 4(2 − 3) = 2ab2 − 9a2 b + 6ab + 10
= 6 · 8 + 4(−1) 5. (5x2 + 4xy − 3y 2 + 2) − (9x2 − 4xy + 2y 2 − 1)
= 48 − 4 = (5x2 + 4xy − 3y 2 + 2) + (−9x2 + 4xy − 2y 2 + 1)
= 44 = (5 − 9)x2 + (4 + 4)xy + (−3 − 2)y 2 + (2 + 1)
= −4x2 + 8xy − 5y 2 + 3
11. Introduction to Polynomials
13. Multiply Polynomials
1. 5 − x6
The term of highest degree is −x6 , so the degree of the 1. (3a2 )(−7a4 ) = [3(−7)](a2 · a4 )
polynomial is 6.
= −21a6
2. x2 y 5 − x7 y + 4
2. (y − 3)(y + 5)
The degree of x2 y 5 is 2 + 5, or 7; the degree of −x7 y is
= y 2 + 5y − 3y − 15 Using FOIL
7 + 1, or 8; the degree of 4 is 0 (4 = 4x0 ). Thus the degree
of the polynomial is 8. = y + 2y − 15
2
Collecting like terms
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 5
3. (x + 6)(x + 3) 3. 3x3 − x2 + 18x − 6
= x2 + 3x + 6x + 18 Using FOIL = x2 (3x − 1) + 6(3x − 1)
2
= x + 9x + 18 Collecting like terms = (3x − 1)(x2 + 6)
4. (2a + 3)(a + 5) 4. t3 + 6t2 − 2t − 12
= 2a2 + 10a + 3a + 15 Using FOIL = t2 (t + 6) − 2(t + 6)
= 2a2 + 13a + 15 Collecting like terms = (t + 6)(t2 − 2)
5. (2x + 3y)(2x + y) 5. w2 − 7w + 10
= 4x2 + 2xy + 6xy + 3y 2 Using FOIL We look for two numbers with a product of 10 and a sum
2
= 4x + 8xy + 3y 2 of −7. By trial, we determine that they are −5 and −2.
w2 − 7w + 10 = (w − 5)(w − 2)
6. (11t − 1)(3t + 4)
= 33t2 + 44t − 3t − 4 Using FOIL 6. t2 + 8t + 15
= 33t2 + 41t − 4 We look for two numbers with a product of 15 and a sum
of 8. By trial, we determine that they are 3 and 5.
t2 + 8t + 15 = (t + 3)(t + 5)
14. Special Products of Binomials
7. 2n2 − 20n − 48 = 2(n2 − 10n − 24)
Now factor n2 − 10n − 24. We look for two numbers with a
1. (x + 3)2 product of −24 and a sum of −10. By trial, we determine
= x2 + 2 · x · 3 + 32 that they are 2 and −12. Then n2 − 10n − 24 =
[(A + B)2 = A2 + 2AB + B 2 ] (n + 2)(n − 12). We must include the common factor, 2,
to have a factorization of the original trinomial.
= x2 + 6x + 9
2n2 − 20n − 48 = 2(n + 2)(n − 12)
2. (5x − 3) 2
8. y 4 − 9y 3 + 14y 2 = y 2 (y 2 − 9y + 14)
= (5x)2 − 2 · 5x · 3 + 32
Now factor y 2 − 9y + 14. Look for two numbers with a
[(A − B)2 = A2 − 2AB + B 2 ] product of 14 and a sum of −9. The numbers are −2 and
= 25x2 − 30x + 9 −7. Then y 2 − 9y + 14 = (y − 2)(y − 7). We must include
the common factor, y 2 , in order to have a factorization of
3. (2x + 3y)2
the original trinomial.
= (2x)2 + 2(2x)(3y) + (3y)2
y 4 − 9y 3 + 14y 2 = y 2 (y − 2)(y − 7)
[(A+B)2 = A2 +2AB +B 2 ]
9. 2n2 + 9n − 56
= 4x2 + 12xy + 9y 2
1. There is no common factor other than 1 or −1.
4. (a − 5b)2
2. The factorization must be of the form
= a2 − 2 · a · 5b + (5b)2 (2n+ )(n+ ).
[(A−B)2 = A2 −2AB +B 2 ] 3. Factor the constant term, −56. The possibilities
= a2 − 10ab + 25b2 are −1 · 56, 1(−56), −2 · 28, 2(−28), −4 · 16, 4(−16),
−7 · 8, and 7(−8). The factors can be written in
5. (n + 6)(n − 6) the opposite order as well: 56(−1), −56 · 1, 28(−2),
= n2 − 62 [(A + B)(A − B) = A2 − B 2 ] −28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7.
= n2 − 36 4. Find a pair of factors for which the sum of the outer
and the inner products is the middle term, 9n. By
6. (3y + 4)(3y − 4)
trial, we determine that the factorization is (2n −
= (3y)2 − 42 [(A + B)(A − B) = A2 − B 2 ] 7)(n + 8).
= 9y 2 − 16
10. 2y 2 + y − 6
1. There is no common factor other than 1 or −1.
15. Factor Polynomials; The FOIL Method 2. The factorization must be of the form
(2y+ )(y+ ).
3. Factor the constant term, −6. The possibilities are
1. 3x + 18 = 3 · x + 3 · 6 = 3(x + 6)
−1 · 6, 1(−6), −2 · 3, and 2(−3). The factors can be
2. 2z 3 − 8z 2 = 2z 2 · z − 2z 2 · 4 = 2z 2 (z − 4) written in the opposite order as well: 6(−1), −6 · 1,
3(−2) and −3 · 2.
Copyright
c 2025 Pearson Education, Inc.
,6 Just-in-Time Review
4. Find a pair of factors for which the sum of the outer 3. 18a2 − 51a + 15
and the inner products is the middle term, y. By 1. Factor out the largest common factor, 3.
trial, we determine that the factorization is
(2y − 3)(y + 2). 18a2 − 51a + 15 = 3(6a2 − 17a + 5)
Now factor 6a2 − 17a + 5.
11. b2 − 6bt + 5t2
2. Multiply the leading coefficient and the constant:
We look for two numbers with a product of 5 and a sum 6(5) = 30.
of −6. By trial, we determine that they are −1 and −5.
3. Try to factor 30 so that the sum of the factors is the
b2 − 6bt + 5t2 = (b − t)(b − 5t) coefficient of the middle term, −17. The factors we
12. x4 − 7x2 − 30 = (x2 )2 − 7x2 − 30 want are −2 and −15.
We look for two numbers with a product of −30 and a sum 4. Split the middle term using the numbers found in
of −7. By trial, we determine that they are 3 and −10. step (3):
x4 − 7x2 − 30 = (x2 + 3)(x2 − 10) −17a = −2a − 15a.
5. Factor by grouping.
6a2 − 17a + 5 = 6a2 − 2a − 15a + 5
16. Factoring Polynomials; The ac-Method
= 2a(3a − 1) − 5(3a − 1)
= (3a − 1)(2a − 5)
1. 8x2 − 6x − 9 Include the largest common factor in the final factoriza-
1. There is no common factor other than 1 or −1. tion.
2. Multiply the leading coefficient and the constant: 18a2 − 51a + 15 = 3(3a − 1)(2a − 5)
8(−9) = −72.
3. Try to factor −72 so that the sum of the factors is
the coefficient of the middle term, −6. The factors 17. Special Factorizations
we want are −12 and 6.
4. Split the middle term using the numbers found in 1. z 2 − 81 = z 2 − 92 = (z + 9)(z − 9)
step (3):
2. 16x2 − 9 = (4x)2 − 32 = (4x + 3)(4x − 3)
−6x = −12x + 6x
5. Factor by grouping. 3. 7pq 4 − 7py 4 = 7p(q 4 − y 4 )
8x2 − 6x − 9 = 8x2 − 12x + 6x − 9 = 7p[(q 2 )2 − (y 2 )2 ]
= 4x(2x − 3) + 3(2x − 3) = 7p(q 2 + y 2 )(q 2 − y 2 )
= (2x − 3)(4x + 3) = 7p(q 2 + y 2 )(q + y)(q − y)
2. 10t2 + 4t − 6 4. x2 + 12x + 36 = x2 + 2 · x · 6 + 62
1. Factor out the largest common factor, 2. = (x + 6)2
10t2 + 4t − 6 = 2(5t2 + 2t − 3) 5. 9z 2 − 12z + 4 = (3z)2 − 2 · 3z · 2 + 22 = (3z − 2)2
Now factor 5t2 + 2t − 3.
6. a3 + 24a2 + 144a
2. Multiply the leading coefficient and the constant:
= a(a2 + 24a + 144)
5(−3) = −15.
= a(a2 + 2 · a · 12 + 122 )
3. Try to factor −15 so that the sum of the factors is
the coefficient of the middle term, 2. The factors we = a(a + 12)2
want are 5 and −3. 7. x3 + 64 = x3 + 43
4. Split the middle term using the numbers found in = (x + 4)(x2 − 4x + 16)
step (3):
8. m3 − 216 = m3 − 63
2t = 5t − 3t.
= (m − 6)(m2 + 6m + 36)
5. Factor by grouping.
5t2 + 2t − 3 = 5t2 + 5t − 3t − 3 9. 3a5 − 24a2 = 3a2 (a3 − 8)
= 5t(t + 1) − 3(t + 1) = 3a2 (a3 − 23 )
= (t + 1)(5t − 3) = 3a2 (a − 2)(a2 + 2a + 4)
Include the largest common factor in the final factoriza- 10. t6 + 1 = (t2 )3 + 13
tion. = (t2 + 1)(t4 − t2 + 1)
10t + 4t − 6 = 2(t + 1)(5t − 3)
2
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 7
2
18. Equation-Solving Principles 2. − x > 6
3
3 3
x < − ·6 Multiplying by − and
1. 7t = 70 2 2
reversing the inequality symbol
t = 10 Dividing by 7 x < −9
The solution is 10. The solution set is (−∞, −9).
2. x − 5 = 7
3. 9x − 1 < 17
x = 12 Adding 5
9x < 18 Adding 1
The solution is 12.
x<2 Dividing by 9
3. 3x + 4 = −8
The solution set is (−∞, 2).
3x = −12 Subtracting 4
x = −4 Dividing by 3 4. −x − 16 ≥ 40
The solution is −4. −x ≥ 56 Adding 6
4. 6x − 15 = 45 x ≤ −56 Multiplying by −1 and
reversing the inequality symbol
6x = 60 Adding 15
The solution set is (−∞, −56].
x = 10 Dividing by 6
The solution is 10.
5. 1
y−6 < 3
5. 7y − 1 = 23 − 5y 3
12y − 1 = 23 Adding 5y 1
y<9 Adding 6
3
12y = 24 Adding 1
y < 27 Multiplying by 3
y=2 Dividing by 12
The solution set is (−∞, 27).
The solution is 2.
6. 3m − 7 = −13 + m 6. 8 − 2w ≤ −14
2m − 7 = −13 Subtracting m −2w ≤ −22 Subtracting 8
2m = −6 Adding 7 w ≥ 11 Dividing by −2 and
m = −3 Dividing by 2 reversing the inequality symbol
The solution is −3. The solution set is [11, ∞).
7. 2(x + 7) = 5x + 14
2x + 14 = 5x + 14 20. The Principle of Zero Products
−3x + 14 = 14 Subtracting 5x
−3x = 0 Subtracting 14 1. 2y 2 + 42y = 0
x=0 2y(y + 21) = 0
The solution is 0. 2y = 0 or y + 21 = 0
8. 5y − (2y − 10) = 25 y = 0 or y = −21
5y − 2y + 10 = 25 The solutions are 0 and −21.
3y + 10 = 25 Collecting like terms 2. (a + 7)(a − 1) = 0
3y = 15 Subtracting 10 a+7 = 0 or a − 1 = 0
y=5 Dividing by 3 a = −7 or a=1
The solution is 5. The solutions are −7 and 1.
3. (5y + 3)(y − 4) = 0
19. Inequality-Solving Principles 5y + 3 = 0 or y − 4 = 0
5y = −3 or y=4
1. p + 25 ≥ −100 3
y = − or y=4
p ≥ −125 Subtracting 25 5
3
The solution set is [−125, ∞). The solutions are − and 4.
5
Copyright
c 2025 Pearson Education, Inc.
, 8 Just-in-Time Review
4. 6x2 + 7x − 5 = 0 3. 6z 2 = 18
(3x + 5)(2x − 1) = 0 z2 = 3
√ √
3x + 5 = 0 or 2x − 1 = 0 z = 3 or z = − 3
√ √ √
3x = −5 or 2x = 1 The solutions are 3 and − 3, or ± 3.
5 1
x = − or x= 4. 3t2 − 15 = 0
3 2
5 1 3t2 = 15
The solutions are − and .
3 2 t2 = 5
√ √
5. t(t − 8) = 0 t= 5 or t = − 5
√ √ √
t = 0 or t − 8 = 0 The solutions are 5 and − 5, or ± 5.
t = 0 or t=8
5. z 2 − 1 = 24
The solutions are 0 and 8.
z 2 = 25
6. x − 8x − 33 = 0
2 √ √
z = 25 or z = − 25
(x + 3)(x − 11) = 0
The solutions are 5 and −5, or ±5.
x+3 = 0 or x − 11 = 0
6. 5x2 − 75 = 0
x = −3 or x = 11
5x2 = 75
The solutions are −3 and 11.
x2 = 15
2
7. x + 13x = 30 √ √
x = 15 or x = − 15
x2 + 13x − 30 = 0 √ √ √
The solutions are 15 and − 15, or ± 15.
(x + 15)(x − 2) = 0
x + 15 = 0 or x − 2 = 0
x = −15 or x=2 22. Simplify Rational Expressions
The solutions are −15 and 2.
3x − 3
8. 12x2 − 7x − 12 = 0 1.
x(x − 1)
(4x + 3)(3x − 4) = 0
The denominator is 0 when the factor x = 0 and also
4x + 3 = 0 or 3x − 4 = 0 when x − 1 = 0, or x = 1. The domain is the set of all real
4x = −3 or 3x = 4 numbers except 0 and 1.
3 4 y+6 y+6
x = − or x= 2. =
4 3 y 2 + 4y − 21 (y + 7)(y − 3)
3 4 The denominator is 0 when y = −7 or y = 3. The domain
The solutions are − and .
4 3 is the set of all real numbers except −7 and 3.
x2 − 4
(x + 2)(x−2) x + 2
21. The Principle of Square Roots
3.
x2 − 4x + 4
=
=
(x − 2)(x−2) x − 2
x2 + 2x − 3 x−1
(x − 1)(x+3)
1. x2 − 36 = 0
4.
x2 − 9
=
(x − 3) = x − 3
(x+3)
x2 = 36
5. x − 6x + 9x = x(x − 6x + 9)
3 2 2
√ √
x = 36 or x = − 36 x3 − 3x2 x2 (x − 3)
x=6 or x = −6 x
/(x−3)(x − 3)
The solutions are 6 and −6, or ±6.
=
/ · x(x−3)
x
x−3
2. 2y 2 − 20 = 0 =
x
2y 2 = 20
6. 6y + 12y − 48 = 6(y + 2y − 8)
2 2
y 2 = 10 3y − 9y + 6
2 3(y 2 − 3y + 2)
√ √
y= 10 or y = − 10
√ √ √
2 · 3/ · (y + 4)(y−2)
The solutions are 10 and − 10, or ± 10.
=
3/(y − 1)(y−2)
2(y + 4)
=
y−1
Copyright
c 2025 Pearson Education, Inc.
Graphs and Models, A Right Triangle Approach, 7th
edition Bittinger
Notes
1- The file is chapter after chapter.
2- We have shown you few pages sample.
3- The file contains all Appendix and Excel sheet
if it exists.
4- We have all what you need, we make update
at every time. There are many new editions
waiting you.
5- If you think you purchased the wrong file You
can contact us at every time, we can replace it
with true one.
Our email:
,Just-in-Time Review
√ √ √ √
4. 6 = 6, so it is true that 6 ≤ 6.
1. Real Numbers
5. −30 is to the left of −25 on the number line, so it is false
that −30 > −25.
2 √ 8 4 16 5 25 16 25
1. Rational numbers: , 6, −2.45, 18.4, −11, 3 27, − , 6. − =− and − = − ; − is to the right of − ,
√ 3 7 5 20 4 20 20 20
0, 16 4 5
so it is true that − > − .
2 8 5 4
2. Rational numbers but not integers: , −2.45, 18.4, −
3 7
√ √ √ √
3. Irrational numbers: 3, 6 26, 7.151551555 . . . , − 35, 5 3 4. Absolute Value
(Although there is a pattern in 7.151551555 . . . , there is
no repeating block of digits.)
√ √ 1. | − 98| = 98 (|a| = −a, if a < 0.)
4. Integers: 6, −11, 3 27, 0, 16
2. |0| = 0 (|a| = a, if a ≥ 0.)
√ √
5. Whole numbers: 6, 3 27, 0, 16
3. |4.7| = 4.7 (|a| = a, if a ≥ 0.)
6. Real numbers: All of them
2 2
4. − = (|a| = −a, if a < 0.)
3 3
2. Properties of Real Numbers 5. | − 7 − 13| = | − 20| = 20, or
|13 − (−7)| = |13 + 7| = |20| = 20
1. −24 + 24 = 0 illustrates the additive inverse property. 6. |2 − 14.6| = | − 12.6| = 12.6, or
2. 7(xy) = (7x)y illustrates the associative property of mul- |14.6 − 2| = |12.6| = 12.6
tiplication.
7. | − 39 − (−28)| = | − 39 + 28| = | − 11| = 11, or
3. 9(r − s) = 9r − 9s illustrates a distributive property. | − 28 − (−39)| = | − 28 + 39| = |11| = 11
4. 11 + z = z + 11 illustrates the commutative property of 3 15 6 15 21 21
addition. 8. − − = − − = − = , or
4 8 8 8 8 8
5. −20 · 1 = −20 illustrates the multiplicative identity prop- 15
− − 3 = 15 + 6 = 21 = 21
erty. 8 4 8 8 8 8
6. 5(x + y) = (x + y)5 illustrates the commutative property
of multiplication.
5. Operations Using Fraction Notation
7. q + 0 = q illustrates the additive identity property.
1 1 3 1 4 3 5 4 15 4 + 15 19
8. 75 · = 1 illustrates the multiplicative inverse property. 1. + = · + · = + = =
75 5 4 5 4 4 5 20 20 20 20
9. (x+y)+w = x+(y+w) illustrates the associative property 3 1 3 1 5 3 5 8 4·2 4
2. + = + · = + = = =
of addition. 10 2 10 2 5 10 10 10 5·2 5
10. 8(a + b) = 8a + 8b illustrates a distributive property. 3 5 3 3 5 4 9 20 29
3. + = · + · = + =
8 6 8 3 6 4 24 24 24
5 5 9 5 27 32
4. +3= +3· = + =
3. Order on the Number Line 9 9 9 9 9 9
7 3 7 3 2 7 6 1
5. − = − · = − =
1. 9 is to the right of −9 on the number line, so it is false 8 4 8 4 2 8 8 8
that 9 < −9. 10 2 10 3 2 11 30 22 8
6. − = · − · = − =
2. −10 is to the left of −1 on the number line, so it is true 11 3 11 3 3 11 33 33 33
that −10 ≤ −1. 3 7 3 14 3 11
7. 2 − =2· − = − =
√ √ √ 7 7 7 7 7 7
3. −5 = − 25, and − 26 is to the left of√− 25, or −5, on
the number line. Thus it is true that − 26 < −5. 5 3 5·3 15 5·3 3
8. · = = = =
8 10 8 · 10 80 5 · 16 16
Copyright
c 2025 Pearson Education, Inc.
,2 Just-in-Time Review
3 5 3·5 15 18. First we write each mixed numeral as a fraction.
9. · = =
4 8 4·8 32 1 1 5 1 5 1 6
1 =1+ =1· + = + =
5 2 5 3 15 3·5 5 5 5 5 5 5 5 5
10. ÷ = · = = = 1 1 4 1 8 1 9
6 3 6 2 12 3·4 4 2 =2+ =2· + = + =
4 4 4 4 4 4 4
3 4 48
11. 12 ÷ = 12 · = = 16 Then we divide.
4 3 3 6 9 6 4 24 8·3 8
÷ = · = = =
5 5 4 5 9 45 15 · 3 15
5 3 5 4 20 10 · 2 10
12. 6 = ÷ = · = = =
3 6 4 6 3 18 9·2 9
4 6. Operations with Real Numbers
1 1 6 1 30 1 31
13. 5 =5+ =5· + = + =
6 6 6 6 6 6 6 1. 8 − (−11) = 8 + 11 = 19
14. We divide.
3 1 3·1 3 1 1 1
2. − · − = = · =1· =
3 10 3 10 · 3 3 10 10 10
9 32 32 5
=3 3. 15 ÷ (−3) = −5
−27 9 9
5 4. −4 − (−1) = −4 + 1 = −3
1 5. 7 · (−50) = −350
15. First we write 4as a fraction.
2 6. −0.5 − 5 = −0.5 + (−5) = −5.5
1 1 2 1 8 1 9
4 =4+ =4· + = + = 7. −3 + 27 = 24
2 2 2 2 2 2 2
Then we add. 8. −400 ÷ −40 = 10
1 2 9 2 9 3 2 2 27 4 31 9. 4.2 · (−3) = −12.6
4 + = + = · + · = + =
2 3 2 3 2 3 3 2 6 6 6
10. −13 − (−33) = −13 + 33 = 20
31
Finally we divide to write as a mixed numeral.
6 11. −60 + 45 = −15
5 1 2 1 2 3 4 1
6 31 12. − = + − = + − =−
31 1 2 3 2 3 6 6 6
=5
−30 6 6 13. −24 ÷ 3 = −8
1
14. −6 + (−16) = −22
16. First we write each mixed numeral as a fraction.
1 5 1 8 1·8 1·2/·4 4
5 5 9 5 18 5 23 15. − ÷ − =− · − = = =
2 =2+ =2· + = + = 2 8 2 5 2·5 2·5
/ 5
9 9 9 9 9 9 9
1 1 3 1 3 1 4
1 =1+ =1· + = + = 7. Interval Notation
3 3 3 3 3 3 3
Then we subtract.
1. This is a closed interval, so we use brackets. Interval no-
23 4 23 4 3 23 12 11 tation is [−5, 5].
− = − · = − =
9 3 9 3 3 9 9 9
2. This is a half-open interval. We use a parenthesis on
11
Finally we divide to write as a mixed numeral. the left and a bracket on the right. Interval notation is
9 (−3, −1].
1
9 11 3. This interval is of unlimited extent in the negative direc-
11 2
=1 tion, and the endpoint −2 is included. Interval notation is
−9 9 9 (−∞, −2].
2
4. This interval is of unlimited extent in the positive direc-
1 tion, and the endpoint 3.8 is not included. Interval nota-
17. First we write 3 as a fraction.
3 tion is (3.8, ∞).
1 1 3 1 9 1 10
3 =3+ =3· + = + = 5. {x|7 < x}, or {x|x > 7}.
3 3 3 3 3 3 3
Then we multiply and simplify. This interval is of unlimited extent in the positive direction
1 9 10 9 90 and the endpoint 7 is not included. Interval notation is
3 · = · = =3 (7, ∞).
3 10 3 10 30
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 3
6. The endpoints −2 and 2 are not included in the interval, 2. Convert 0.000786 to scientific notation.
so we use parentheses. Interval notation is (−2, 2). We want the decimal point to be positioned between the
7. The endpoints −4 and 5 are not included in the interval, 7 and the 8, so we move it 4 places to the right. Since
so we use parentheses. Interval notation is (−4, 5). 0.000786 is between 0 and 1, the exponent must be nega-
tive.
8. The interval is of unlimited extent in the positive direc- 0.000786 = 7.86 × 10−4
tion, and the endpoint 1.7 is included. Internal notation
is [1.7, ∞). 3. Convert 0.0000000023 to scientific notation.
9. The endpoint −5 is not included in the interval, so we use a We want the decimal point to be positioned between the
parenthesis before −5. The endpoint −2 is included in the 2 and the 3, so we move it 9 places to the right. Since
interval, so we use a bracket after −2. Interval notation is 0.0000000023 is between 0 and 1, the exponent must be
(−5, −2]. negative.
0.0000000023 = 2.3 × 10−9
10. This interval is of unlimited
√ extent in the negative direc-
tion, and the√endpoint 5 is not included. Interval nota- 4. Convert 8,927,000,000 to scientific notation.
tion is (−∞, 5). We want the decimal point to be positioned between the
8 and the 9, so we move it 9 places to the left. Since
8,927,000,000 is greater than 10, the exponent must be
8. Integers as Exponents positive.
8, 927, 000, 000 = 8.927 × 109
1 1 5. Convert 4.3 × 10−8 to decimal notation.
1. 3−6 = Using a−m =
36 am
The exponent is negative, so the number is between 0 and
1 −m 1 1. We move the decimal point 8 places to the left.
2. = (0.2)5 Using a = m
(0.2)−5 a
4.3 × 10−8 = 0.000000043
w−4 z9 a−m bn
3. −9
= 4 Using −n = m 6. Convert 5.17 × 106 to decimal notation.
z w b a
2 The exponent is positive, so the number is greater than
z z2 10. We move the decimal point 6 places to the right.
4. = 2 Raising a quotient to a power
y y 5.17 × 106 = 5, 170, 000
5. 1000 = 1 Using a0 = 1, a = 0 7. Convert 6.203 × 1011 to decimal notation.
a5 The exponent is positive, so the number is greater than
6. = a5−(−3) = a5+3 = a8 Using the quotient rule
a−3 10. We move the decimal point 11 places to the right.
7. (2xy 3 )(−3x−5 y) = 2(−3)x · x−5 · y 3 · y 6.203 × 1011 = 620, 300, 000, 000
= −6x1+(−5) y 3+1 8. Convert 2.94 × 10−5 to scientific notation.
4
6y
= −6x−4 y 4 , or − The exponent is negative, so the number is between 0 and
x4 1. We move the decimal point 5 places to the left.
1 2.94 × 10−5 = 0.0000294
8. x−4 · x−7 = x−4+(−7) = x−11 , or
x11
1
9. (mn)−6 = m−6 n−6 , or 10. Order of Operations
m6 n6
1
10. (t−5 )4 = t−5·4 = t−20 , or
t20 1. 3 + 18 ÷ 6 − 3 = 3 + 3 − 3 Dividing
= 6−3=3 Adding and subtracting
9. Scientific Notation 2. = 5 · 3 + 8 · 32 + 4(6 − 2)
= 5 · 3 + 8 · 32 + 4 · 4 Working inside parentheses
1. Convert 18,500,000 to scientific notation. = 5·3+8·9+4·4 Evaluating 32
We want the decimal point to be positioned between the = 15 + 72 + 16 Multiplying
1 and the 8, so we move it 7 places to the left. Since = 87 + 16 Adding in order
18,500,000 is greater than 10, the exponent must be posi-
tive. = 103 from left to right
18, 500, 000 = 1.85 × 107
Copyright
c 2025 Pearson Education, Inc.
,4 Just-in-Time Review
3. 5[3 − 8 · 32 + 4 · 6 − 2] 3. 2a4 − 3 + a2
= 5[3 − 8 · 9 + 4 · 6 − 2] The term of highest degree is 2a4 , so the degree of the
= 5[3 − 72 + 24 − 2] polynomial is 4.
= 5[−69 + 24 − 2] 4. −41 = −41x0 , so the degree of the polynomial is 0.
= 5[−45 − 2]
5. 4x − x3 + 0.1x8 − 2x5
= 5[−47]
The term of highest degree is 0.1x8 , so the degree of the
= −235 polynomial is 8.
4. 16 ÷ 4 · 4 ÷ 2 · 256 6. x − 3 has two terms. It is a binomial.
= 4 · 4 ÷ 2 · 256 Multiplying and dividing
7. 14y 5 has one term. It is a monomial.
in order from left to right
= 16 ÷ 2 · 256 1
8. 2y − y 2 + 8 has three terms. It is a trinomial.
= 8 · 256 4
= 2048
5. 26 · 2−3 ÷ 210 ÷ 2−8 12. Add and Subtract Polynomials
−8
= 2 ÷2
3 10
÷2
−7 −8
=2 ÷2 1. (8y − 1) − (3 − y)
=2 = (8y − 1) + (−3 + y)
6. 4(8 − 6)2 − 4 · 3 + 2 · 8 = (8 + 1)y + (−1 − 3)
31 + 190 = 9y − 4
4 · 22 − 4 · 3 + 2 · 8
= Calculating in the 2. (3x2 − 2x − x3 + 2) − (5x2 − 8x − x3 + 4)
3+1
numerator and in = (3x2 − 2x − x3 + 2) + (−5x2 + 8x + x3 − 4)
the denominator
4·4−4·3+2·8 = (3 − 5)x2 + (−2 + 8)x + (−1 + 1)x3 + (2 − 4)
= = −2x2 + 6x − 2
4
16 − 12 + 16
= 3. (2x + 3y + z − 7) + (4x − 2y − z + 8)+
4
4 + 16 (−3x + y − 2z − 4)
= = (2 + 4 − 3)x + (3 − 2 + 1)y + (1 − 1 − 2)z+
4
20 (−7 + 8 − 4)
=
4 = 3x + 2y − 2z − 3
= 5
4. (3ab2 − 4a2 b − 2ab + 6)+
7. 64 ÷ [(−4) ÷ (−2)] = 64 ÷ 2 = 32
(−ab2 − 5a2 b + 8ab + 4)
8. 6[9 − (3 − 2)] + 4(2 − 3) = (3 − 1)ab2 + (−4 − 5)a2 b + (−2 + 8)ab + (6 + 4)
= 6[9 − 1] + 4(2 − 3) = 2ab2 − 9a2 b + 6ab + 10
= 6 · 8 + 4(−1) 5. (5x2 + 4xy − 3y 2 + 2) − (9x2 − 4xy + 2y 2 − 1)
= 48 − 4 = (5x2 + 4xy − 3y 2 + 2) + (−9x2 + 4xy − 2y 2 + 1)
= 44 = (5 − 9)x2 + (4 + 4)xy + (−3 − 2)y 2 + (2 + 1)
= −4x2 + 8xy − 5y 2 + 3
11. Introduction to Polynomials
13. Multiply Polynomials
1. 5 − x6
The term of highest degree is −x6 , so the degree of the 1. (3a2 )(−7a4 ) = [3(−7)](a2 · a4 )
polynomial is 6.
= −21a6
2. x2 y 5 − x7 y + 4
2. (y − 3)(y + 5)
The degree of x2 y 5 is 2 + 5, or 7; the degree of −x7 y is
= y 2 + 5y − 3y − 15 Using FOIL
7 + 1, or 8; the degree of 4 is 0 (4 = 4x0 ). Thus the degree
of the polynomial is 8. = y + 2y − 15
2
Collecting like terms
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 5
3. (x + 6)(x + 3) 3. 3x3 − x2 + 18x − 6
= x2 + 3x + 6x + 18 Using FOIL = x2 (3x − 1) + 6(3x − 1)
2
= x + 9x + 18 Collecting like terms = (3x − 1)(x2 + 6)
4. (2a + 3)(a + 5) 4. t3 + 6t2 − 2t − 12
= 2a2 + 10a + 3a + 15 Using FOIL = t2 (t + 6) − 2(t + 6)
= 2a2 + 13a + 15 Collecting like terms = (t + 6)(t2 − 2)
5. (2x + 3y)(2x + y) 5. w2 − 7w + 10
= 4x2 + 2xy + 6xy + 3y 2 Using FOIL We look for two numbers with a product of 10 and a sum
2
= 4x + 8xy + 3y 2 of −7. By trial, we determine that they are −5 and −2.
w2 − 7w + 10 = (w − 5)(w − 2)
6. (11t − 1)(3t + 4)
= 33t2 + 44t − 3t − 4 Using FOIL 6. t2 + 8t + 15
= 33t2 + 41t − 4 We look for two numbers with a product of 15 and a sum
of 8. By trial, we determine that they are 3 and 5.
t2 + 8t + 15 = (t + 3)(t + 5)
14. Special Products of Binomials
7. 2n2 − 20n − 48 = 2(n2 − 10n − 24)
Now factor n2 − 10n − 24. We look for two numbers with a
1. (x + 3)2 product of −24 and a sum of −10. By trial, we determine
= x2 + 2 · x · 3 + 32 that they are 2 and −12. Then n2 − 10n − 24 =
[(A + B)2 = A2 + 2AB + B 2 ] (n + 2)(n − 12). We must include the common factor, 2,
to have a factorization of the original trinomial.
= x2 + 6x + 9
2n2 − 20n − 48 = 2(n + 2)(n − 12)
2. (5x − 3) 2
8. y 4 − 9y 3 + 14y 2 = y 2 (y 2 − 9y + 14)
= (5x)2 − 2 · 5x · 3 + 32
Now factor y 2 − 9y + 14. Look for two numbers with a
[(A − B)2 = A2 − 2AB + B 2 ] product of 14 and a sum of −9. The numbers are −2 and
= 25x2 − 30x + 9 −7. Then y 2 − 9y + 14 = (y − 2)(y − 7). We must include
the common factor, y 2 , in order to have a factorization of
3. (2x + 3y)2
the original trinomial.
= (2x)2 + 2(2x)(3y) + (3y)2
y 4 − 9y 3 + 14y 2 = y 2 (y − 2)(y − 7)
[(A+B)2 = A2 +2AB +B 2 ]
9. 2n2 + 9n − 56
= 4x2 + 12xy + 9y 2
1. There is no common factor other than 1 or −1.
4. (a − 5b)2
2. The factorization must be of the form
= a2 − 2 · a · 5b + (5b)2 (2n+ )(n+ ).
[(A−B)2 = A2 −2AB +B 2 ] 3. Factor the constant term, −56. The possibilities
= a2 − 10ab + 25b2 are −1 · 56, 1(−56), −2 · 28, 2(−28), −4 · 16, 4(−16),
−7 · 8, and 7(−8). The factors can be written in
5. (n + 6)(n − 6) the opposite order as well: 56(−1), −56 · 1, 28(−2),
= n2 − 62 [(A + B)(A − B) = A2 − B 2 ] −28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7.
= n2 − 36 4. Find a pair of factors for which the sum of the outer
and the inner products is the middle term, 9n. By
6. (3y + 4)(3y − 4)
trial, we determine that the factorization is (2n −
= (3y)2 − 42 [(A + B)(A − B) = A2 − B 2 ] 7)(n + 8).
= 9y 2 − 16
10. 2y 2 + y − 6
1. There is no common factor other than 1 or −1.
15. Factor Polynomials; The FOIL Method 2. The factorization must be of the form
(2y+ )(y+ ).
3. Factor the constant term, −6. The possibilities are
1. 3x + 18 = 3 · x + 3 · 6 = 3(x + 6)
−1 · 6, 1(−6), −2 · 3, and 2(−3). The factors can be
2. 2z 3 − 8z 2 = 2z 2 · z − 2z 2 · 4 = 2z 2 (z − 4) written in the opposite order as well: 6(−1), −6 · 1,
3(−2) and −3 · 2.
Copyright
c 2025 Pearson Education, Inc.
,6 Just-in-Time Review
4. Find a pair of factors for which the sum of the outer 3. 18a2 − 51a + 15
and the inner products is the middle term, y. By 1. Factor out the largest common factor, 3.
trial, we determine that the factorization is
(2y − 3)(y + 2). 18a2 − 51a + 15 = 3(6a2 − 17a + 5)
Now factor 6a2 − 17a + 5.
11. b2 − 6bt + 5t2
2. Multiply the leading coefficient and the constant:
We look for two numbers with a product of 5 and a sum 6(5) = 30.
of −6. By trial, we determine that they are −1 and −5.
3. Try to factor 30 so that the sum of the factors is the
b2 − 6bt + 5t2 = (b − t)(b − 5t) coefficient of the middle term, −17. The factors we
12. x4 − 7x2 − 30 = (x2 )2 − 7x2 − 30 want are −2 and −15.
We look for two numbers with a product of −30 and a sum 4. Split the middle term using the numbers found in
of −7. By trial, we determine that they are 3 and −10. step (3):
x4 − 7x2 − 30 = (x2 + 3)(x2 − 10) −17a = −2a − 15a.
5. Factor by grouping.
6a2 − 17a + 5 = 6a2 − 2a − 15a + 5
16. Factoring Polynomials; The ac-Method
= 2a(3a − 1) − 5(3a − 1)
= (3a − 1)(2a − 5)
1. 8x2 − 6x − 9 Include the largest common factor in the final factoriza-
1. There is no common factor other than 1 or −1. tion.
2. Multiply the leading coefficient and the constant: 18a2 − 51a + 15 = 3(3a − 1)(2a − 5)
8(−9) = −72.
3. Try to factor −72 so that the sum of the factors is
the coefficient of the middle term, −6. The factors 17. Special Factorizations
we want are −12 and 6.
4. Split the middle term using the numbers found in 1. z 2 − 81 = z 2 − 92 = (z + 9)(z − 9)
step (3):
2. 16x2 − 9 = (4x)2 − 32 = (4x + 3)(4x − 3)
−6x = −12x + 6x
5. Factor by grouping. 3. 7pq 4 − 7py 4 = 7p(q 4 − y 4 )
8x2 − 6x − 9 = 8x2 − 12x + 6x − 9 = 7p[(q 2 )2 − (y 2 )2 ]
= 4x(2x − 3) + 3(2x − 3) = 7p(q 2 + y 2 )(q 2 − y 2 )
= (2x − 3)(4x + 3) = 7p(q 2 + y 2 )(q + y)(q − y)
2. 10t2 + 4t − 6 4. x2 + 12x + 36 = x2 + 2 · x · 6 + 62
1. Factor out the largest common factor, 2. = (x + 6)2
10t2 + 4t − 6 = 2(5t2 + 2t − 3) 5. 9z 2 − 12z + 4 = (3z)2 − 2 · 3z · 2 + 22 = (3z − 2)2
Now factor 5t2 + 2t − 3.
6. a3 + 24a2 + 144a
2. Multiply the leading coefficient and the constant:
= a(a2 + 24a + 144)
5(−3) = −15.
= a(a2 + 2 · a · 12 + 122 )
3. Try to factor −15 so that the sum of the factors is
the coefficient of the middle term, 2. The factors we = a(a + 12)2
want are 5 and −3. 7. x3 + 64 = x3 + 43
4. Split the middle term using the numbers found in = (x + 4)(x2 − 4x + 16)
step (3):
8. m3 − 216 = m3 − 63
2t = 5t − 3t.
= (m − 6)(m2 + 6m + 36)
5. Factor by grouping.
5t2 + 2t − 3 = 5t2 + 5t − 3t − 3 9. 3a5 − 24a2 = 3a2 (a3 − 8)
= 5t(t + 1) − 3(t + 1) = 3a2 (a3 − 23 )
= (t + 1)(5t − 3) = 3a2 (a − 2)(a2 + 2a + 4)
Include the largest common factor in the final factoriza- 10. t6 + 1 = (t2 )3 + 13
tion. = (t2 + 1)(t4 − t2 + 1)
10t + 4t − 6 = 2(t + 1)(5t − 3)
2
Copyright
c 2025 Pearson Education, Inc.
,Just-in-Time Review 7
2
18. Equation-Solving Principles 2. − x > 6
3
3 3
x < − ·6 Multiplying by − and
1. 7t = 70 2 2
reversing the inequality symbol
t = 10 Dividing by 7 x < −9
The solution is 10. The solution set is (−∞, −9).
2. x − 5 = 7
3. 9x − 1 < 17
x = 12 Adding 5
9x < 18 Adding 1
The solution is 12.
x<2 Dividing by 9
3. 3x + 4 = −8
The solution set is (−∞, 2).
3x = −12 Subtracting 4
x = −4 Dividing by 3 4. −x − 16 ≥ 40
The solution is −4. −x ≥ 56 Adding 6
4. 6x − 15 = 45 x ≤ −56 Multiplying by −1 and
reversing the inequality symbol
6x = 60 Adding 15
The solution set is (−∞, −56].
x = 10 Dividing by 6
The solution is 10.
5. 1
y−6 < 3
5. 7y − 1 = 23 − 5y 3
12y − 1 = 23 Adding 5y 1
y<9 Adding 6
3
12y = 24 Adding 1
y < 27 Multiplying by 3
y=2 Dividing by 12
The solution set is (−∞, 27).
The solution is 2.
6. 3m − 7 = −13 + m 6. 8 − 2w ≤ −14
2m − 7 = −13 Subtracting m −2w ≤ −22 Subtracting 8
2m = −6 Adding 7 w ≥ 11 Dividing by −2 and
m = −3 Dividing by 2 reversing the inequality symbol
The solution is −3. The solution set is [11, ∞).
7. 2(x + 7) = 5x + 14
2x + 14 = 5x + 14 20. The Principle of Zero Products
−3x + 14 = 14 Subtracting 5x
−3x = 0 Subtracting 14 1. 2y 2 + 42y = 0
x=0 2y(y + 21) = 0
The solution is 0. 2y = 0 or y + 21 = 0
8. 5y − (2y − 10) = 25 y = 0 or y = −21
5y − 2y + 10 = 25 The solutions are 0 and −21.
3y + 10 = 25 Collecting like terms 2. (a + 7)(a − 1) = 0
3y = 15 Subtracting 10 a+7 = 0 or a − 1 = 0
y=5 Dividing by 3 a = −7 or a=1
The solution is 5. The solutions are −7 and 1.
3. (5y + 3)(y − 4) = 0
19. Inequality-Solving Principles 5y + 3 = 0 or y − 4 = 0
5y = −3 or y=4
1. p + 25 ≥ −100 3
y = − or y=4
p ≥ −125 Subtracting 25 5
3
The solution set is [−125, ∞). The solutions are − and 4.
5
Copyright
c 2025 Pearson Education, Inc.
, 8 Just-in-Time Review
4. 6x2 + 7x − 5 = 0 3. 6z 2 = 18
(3x + 5)(2x − 1) = 0 z2 = 3
√ √
3x + 5 = 0 or 2x − 1 = 0 z = 3 or z = − 3
√ √ √
3x = −5 or 2x = 1 The solutions are 3 and − 3, or ± 3.
5 1
x = − or x= 4. 3t2 − 15 = 0
3 2
5 1 3t2 = 15
The solutions are − and .
3 2 t2 = 5
√ √
5. t(t − 8) = 0 t= 5 or t = − 5
√ √ √
t = 0 or t − 8 = 0 The solutions are 5 and − 5, or ± 5.
t = 0 or t=8
5. z 2 − 1 = 24
The solutions are 0 and 8.
z 2 = 25
6. x − 8x − 33 = 0
2 √ √
z = 25 or z = − 25
(x + 3)(x − 11) = 0
The solutions are 5 and −5, or ±5.
x+3 = 0 or x − 11 = 0
6. 5x2 − 75 = 0
x = −3 or x = 11
5x2 = 75
The solutions are −3 and 11.
x2 = 15
2
7. x + 13x = 30 √ √
x = 15 or x = − 15
x2 + 13x − 30 = 0 √ √ √
The solutions are 15 and − 15, or ± 15.
(x + 15)(x − 2) = 0
x + 15 = 0 or x − 2 = 0
x = −15 or x=2 22. Simplify Rational Expressions
The solutions are −15 and 2.
3x − 3
8. 12x2 − 7x − 12 = 0 1.
x(x − 1)
(4x + 3)(3x − 4) = 0
The denominator is 0 when the factor x = 0 and also
4x + 3 = 0 or 3x − 4 = 0 when x − 1 = 0, or x = 1. The domain is the set of all real
4x = −3 or 3x = 4 numbers except 0 and 1.
3 4 y+6 y+6
x = − or x= 2. =
4 3 y 2 + 4y − 21 (y + 7)(y − 3)
3 4 The denominator is 0 when y = −7 or y = 3. The domain
The solutions are − and .
4 3 is the set of all real numbers except −7 and 3.
x2 − 4
(x + 2)(x−2) x + 2
21. The Principle of Square Roots
3.
x2 − 4x + 4
=
=
(x − 2)(x−2) x − 2
x2 + 2x − 3 x−1
(x − 1)(x+3)
1. x2 − 36 = 0
4.
x2 − 9
=
(x − 3) = x − 3
(x+3)
x2 = 36
5. x − 6x + 9x = x(x − 6x + 9)
3 2 2
√ √
x = 36 or x = − 36 x3 − 3x2 x2 (x − 3)
x=6 or x = −6 x
/(x−3)(x − 3)
The solutions are 6 and −6, or ±6.
=
/ · x(x−3)
x
x−3
2. 2y 2 − 20 = 0 =
x
2y 2 = 20
6. 6y + 12y − 48 = 6(y + 2y − 8)
2 2
y 2 = 10 3y − 9y + 6
2 3(y 2 − 3y + 2)
√ √
y= 10 or y = − 10
√ √ √
2 · 3/ · (y + 4)(y−2)
The solutions are 10 and − 10, or ± 10.
=
3/(y − 1)(y−2)
2(y + 4)
=
y−1
Copyright
c 2025 Pearson Education, Inc.
