Algebra: Chapter 9

Front 1

x 2 + 16x + 63.

Back 1

Factors of 63 Sum of Factors
1, 63 64
3, 21 24
7, 9 16 The correct factors are 7 and 9.
x 2 + 16x + 63 = (x + m)(x + n) Write the pattern.
= (x + 7)(x + 9) m = 7 and n = 9
Check: You can check this result by multiplying the two factors.
F O I L
(x + 7)(x + 9) = x 2 + 9x + 7x + 63 FOIL method
= x 2 + 16x + 63 Simplify.

Front 2

x 2 - 11x + 24.

Back 2

Factors of 24 Sum of Factors
-1, -24 -25
-2, -12 -14
-3, -8 -11 The correct factors are –3 and –8.
-4, -6 -10
x 2 - 11x + 24 = (x + m)(x + n) Write the pattern.
= (x - 3)(x - 8) m = -3 and n = -8

Front 3

x 2 + 2x – 15.

Back 3

Factors of –15 Sum of Factors
-1, 15 14
1, -15 -14
-3, 5 2 The correct factors are –3 and 5.
3, -5 -2
x 2 + 2x - 15 = (x + m)(x + n) Write the pattern.
= (x - 3)(x + 5) m = -3 and n = 5

Front 4

x 2 – 4x – 21

Back 4

Factors of –21 Sum of Factors
-1, 21 20
1, -21 -20
-3, 7 4
3, -7 -4 The correct factors are 3 and –7.
x 2 – 4x – 21 = (x + m)(x + n) Write the pattern.
= (x + 3)(x – 7) m = 3 and n = -7

Front 5

x 2 – 8x + 7 = 0

Back 5

x 2 – 8x + 7 = 0 Original equation
(x – 1)(x – 7) = 0 Factor.
x – 1 = 0 or x – 7 = 0 Zero Product Property
x = 1 x = 7 Solve each equation.
The solution set is {1, 7}

Front 6

4x 2 + 8x – 5.

Back 6

Factors of -20 Sum of Factors
-1, 20 19
1, -20 -19
-2, 10 8 The correct factors are –2 and 10.
2, -10 -8
-4, 5 1
4, -5 -1
4x 2 + 8x – 5 = 4x 2 + mx + nx – 5 Write the pattern.
= 4x 2 + -2x + 10x – 5 m = -2 and n = 10
= (4x 2 + -2x) + (10x – 5) Group terms with common factors.
= 2x(2x – 1) + 5(2x – 1) Factor the GCF from each grouping.
= (2x – 1)(2x + 5) 2x – 1 is the common factor.

Front 7

3x 2 + 11x + 10.

Back 7

Factors of 30 Sum of Factors
1, 30 31
2, 15 17
3, 10 13
5, 6 11 The correct factors are 5 and 6.
3x 2 + 11x + 10 = 3x 2 + mx + nx + 10 Write the pattern.
= 3x 2 + 5x + 6x + 10 m = 5 and n = 6
= (3x 2 + 5x) + (6x + 10) Group terms with common factors.
= x(3x + 5) + 2(3x + 5) Factor the GCF from each grouping.
= (3x + 5) (x + 2) Factor out the common factor 3x + 5.

Front 8

36x 2 + 6x – 12.

Back 8

Factors of –12 Sum of Factors
-1, 12 11
1, -12 -11
-2, 6 4
2, -6 -4
-3, 4 1 The correct factors are -3 and 4.
3, -4 -1
6x 2 + x – 2 = 6x 2 + mx + nx – 2 Write the pattern.
= 6x 2 – 3x + 4x – 2 m = -3 and n = 4
= (6x 2 – 3x) + (4x – 2) Group terms with common factors.
= 3x(2x – 1) + 2(2x – 1) Factor the GCF from each grouping.
= (2x – 1)(3x + 2) Factor out the common factor 2x – 1.
Thus the complete factorization is 36x 2 + 6x – 12 = 6(2x – 1)(3x + 2).

Front 9

3x 2 – x + 1.

Back 9

Factors of 3 Sum of Factors
-1, -3 -4
There are no factors whose sum is -1. Therefore, 3x 2 – x + 1 cannot be factored using integers. Thus,
3x 2 – x + 1 is a prime polynomial.

Front 10

Solve k 2 +
3
8 k = 1.

Back 10

k 2 +
3
8 k = 1 Original equation.
3(k 2 +
3
8 k )= 3(1) Eliminate fractions by multiplying each side by 3.
3k 2 + 8k = 3 Distributive Property
3k 2 + 8k – 3 = 0 Rewrite so that one side equals 0.
(3k – 1)(k + 3) = 0 Factor the left side.
3k – 1 = 0 or k + 3 = 0 Zero Product Property
3k = 1 k = -3 Solve each equation.
k =
3
1
The solution set is {-3,
3
1
}.

Front 11

a 2 – 64

Back 11

a 2 – 64 = a 2 – 8 2 Write in the form a 2 – b 2 .
= (a + 8)(a – 8) Factor the difference of squares.

Front 12

16m4 - n 2

Back 12

16m4 – n 2 = (4m2 ) 2 – (n) 2 16m4 = 4m2 ⋅ 4m2 and n 2 = n ⋅ n
= (4m2 + n)(4m2 – n) Factor the difference of squares.

Front 13

5x 2 y – 500y.

Back 13

5x 2 y – 500y = 5y(x 2 – 100) The GCF of 5x 2 y and 500y is 5y.
= 5y(x 2 – 10 2 ) 100 = 10 2
= 5y(x + 10)(x – 10) Factor the difference of squares.

Front 14

16x 5 – 625x.

Back 14

16x 5 – 625x Original polynomial
= x(16x 4 – 625) The GCF of 16x 5 and 625x is x.
= x[(4x 2 ) 2 – 25 2 ] 16x 4 = 4x 2 ⋅ x 2 and 625 = 25 ⋅ 25
= x(4x 2 + 25)( 4x 2 – 25) Factor the difference of squares.
= x(4x 2 + 25)[(2x) 2 – 5 2 ] 4x 2 = 2x ⋅ 2x and 25 = 5 ⋅ 5
= x(4x 2 + 25)(2x + 5)(2x – 5) Factor the difference of squares.

Front 15

a 4 – 5a 3 – 4a 2 + 20a.

Back 15

a 4 – 5a 3 – 4a 2 + 20a Original polynomial
= a(a 3 – 5a 2 – 4a + 20) Factor out the GCF.
= a[(a 3 – 5a 2 ) +(-4a + 20)] Group terms with common factors.
= a[a 2 (a – 5) +4(-a + 5)] Factor each grouping.
= a[a 2 (a – 5) +4(-1)(a – 5)] (-a + 5) = -1(a – 5)
= a[a 2 (a – 5) – 4(a – 5)] Simplify.
= a[(a – 5) (a 2 – 4)] a – 5 is the common factor.
= a[(a – 5) (a 2 – 2 2 )] a 2 = a ⋅ a and 4 = 2 ⋅ 2
= a(a – 5)(a + 2)(a – 2) Factor the difference of squares.

Front 16

a 3 + 2a 2 – a = 2

Back 16

a 3 + 2a 2 – a = 2 Original equation
a 3 + 2a 2 – a – 2 = 0 Subtract 2 from each side.
(a 3 + 2a 2 ) + (- a – 2) = 0 Group terms with a common factor.
a 2 (a+ 2) + -1( a + 2) = 0 Factor each grouping.
(a + 2)( a 2 – 1) = 0 a + 2 is the common factor.
(a + 2)(a + 1)(a – 1) = 0 a 2 = a ⋅ a and 1 = 1 ⋅ 1
Applying the Zero Product Property, set each factor equal to 0 and solve the resulting three equations.
a + 2 = 0 or a + 1 = 0 or a – 1 = 0
a = -2 a = -1 a = 1
The solution set is {-2, -1, 1}. Check each solution in the original equation.

Front 17

4 b 2 – 1 = 0

Back 17

4 b 2 – 1 = 0 Original equation
(2b) 2 – 1 2 = 0 4 b 2 = 2b ⋅ 2b and 1 = 1 ⋅ 1
(2b + 1)(2b – 1) = 0 Factor the difference of squares.
2b + 1 = 0 or 2b – 1 = 0 Zero Product Property
2b = -1 2b = 1 Solve each equation.
b = -
2
1 b =
2
1
The solution set is {-
2
1
,
2
1
}. Check each solution in the original equation.

Front 18

4x 2 + 12x + 9

Back 18

1. Is the first term a perfect square? Yes, 4x 2 = (2x) 2 .
2. Is the last term a perfect square? Yes, 9 = (3) 2 .
3. Is the middle term equal to 2(2x)(3)? Yes, 12x = 2(2x)(3).
4x 2 + 12x + 9 is a perfect square trinomial.
4x 2 + 12x + 9 = (2x) 2 + 2(2x)(3) + (3) 2 Write as a 2 + 2ab + b 2 .
= (2x + 3) 2 Factor using pattern.

Front 19

4a 2 - 8a + 16

Back 19

1. Is the first term a perfect square? Yes, 4a 2 = (2a) 2 .
2. Is the last term a perfect square? Yes, 16 = (4) 2 .
3. Is the middle term equal to 2(2a)(4)? No, 8a ≠ 2(2a)(4).
4a 2 - 8a + 16 is not a perfect square trinomial.

Front 20

3x 2 – 12x + 12

Back 20

This polynomial has a GCF of 3. First, factor out the GCF and you are left with 3(x 2 – 4x + 4). The
resulting trinomial has the first term as a perfect square x 2 = (x) 2 , the last term is also a perfect
square 4 = 2 2 , and the middle term is equal to 2(x)(2) or 4x. Therefore, the polynomial is a perfect
square trinomial.
3x 2 – 12x + 12
= 3(x 2 – 4x + 4) 3 is the GCF.
= 3[(x) 2 – 2(x)(2) + (2) 2 ] Write as a 2 – 2ab + b 2 .
= 3(x – 2) 2 a = x and b = 2.

Front 21

2x 3 – x 2 - 15x

Back 21

2x 3 – x 2 - 15x
= x (2x 2 - x – 15) x is the GCF.
= x(2x 2 + mx + nx – 15) Write the pattern.
= x(2x 2 + 5x + -6x – 15) m = 5 and n = -6
= x[(2x 2 + 5x) + (-6x – 15)] Group terms with common factors.
= x[x(2x + 5) + -3(2x + 5)] Factor out the GCF from each grouping.
= x(2x + 5)(x – 3) 2x + 5 is the common factor.

Front 22

16x 2 + 8x + 1 = 0.

Back 22

16x 2 + 8x + 1 = 0 Original equation
(4x) 2 + 2(4x)(1) + (1) 2 = 0 Recognize 16x 2 – 8x + 1 as a perfect square trinomial.
(4x + 1) 2 = 0 Factor the perfect square trinomial.
4x + 1 = 0 Set repeated factor equal to zero.
4x = -1 Solve for x.
x = -
4
1
Thus, the solution set is {-
4
1
}. Check this solution in the original equation.

Front 23

(x – 2) 2 =
9
4

Back 23

(x – 2) 2 =
9
4
Original equation
x – 2 =
9
± 4 Square Root Property
x – 2 = ±
3
2
9
4
=
3
2 ⋅
3
2
x = 2 ±
3
2
Add 2 to each side.
x = 2 +
3
2
or x = 2 –
3
2
Separate into two equations.
=
3
8
=
3
4
Simplify.
The solution set is {
3
4
,
3
8
}. Check each solution in the original equation.

Front 24

x 2 +
2
1 x +
16
1 = 36

Back 24

x 2 +
2
1 x +
16
1
= 36 Original equation
(x) 2 + 2(x)(
4
1
) + (
4
1
) 2 = 36 Recognize perfect square trinomial.
(x + 1
4
) 2 = 36 Factor perfect square trinomial.
x + 1
4
= ± 36 Square Root Property
x + 1
4
= ± 6 36 = 6 ⋅ 6
x = - 1
4
± 6 Subtract
2
1
from each side.
x = - 1
4
+6 or x = - 1
4
– 6 Separate into two equations.
= 23
4
= - 25
4
− Simplify.
The solution set is { 25
4
− , 23
4
}. Check each solution in the original equation.

Front 25

(x + 1) 2 = 10

Back 25

(x + 1) 2 = 10 Original equation
x + 1 = ± 10 Square Root Property
x = -1 ± 10 Subtract 1 from each side.
Since 10 is not a perfect square, the solution set is {-1 ± 10 }. Using a calculator, the approximate
solutions are -1 + 10 or about 2.16 and -1 - 10 or about –4.16.