Transcript Lesson 8.2b - Crestwood Local Schools

Factoring by Grouping
ALGEBRA 1 LESSON 9-8
(For help, go to Lessons 9-2 and 9-3.)
Find the GCF of the terms of each polynomial.
1. 6y2 + 12y – 4
2. 9r3 + 15r2 + 21r
3. 30h3 – 25h2 – 40h
4. 16m3 – 12m2 – 36m
Find each product.
5. (v + 3)(v2 + 5)
6. (2q2 – 4)(q – 5)
7. (2t – 5)(3t + 4)
8. (4x – 1)(x2 + 2x + 3)
Check Skills You’ll Need
9-8
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Solutions
1. 6y2 + 12y – 4
6y2 = 2 • 3 • y • y;
12y = 2 • 2 • 3 • y; 4 = 2 • 2;
GCF = 2
3. 30h3 – 25h2 – 40h
30h3 = 2 • 3 • 5 • h • h • h;
25h2 = 5 • 5 • h • h;
40h = 2 • 2 • 2 • 5 • h;
GCF = 5h
2. 9r3 + 15r2 + 21r
9r3 = 3 • 3 • r • r • r;
15r2 = 3 • 5 • r • r; 21r = 3 • 7 • r;
GCF = 3r
4. 16m3 – 12m2 – 36m
16m3 = 2 • 2 • 2 • 2 • m • m • m;
12m2 = 2 • 2 • 3 • m • m;
36m = 2 • 2 • 3 • 3 • m;
GCF = 2 • 2 • m = 4m
5. (v + 3)(v2 + 5) = (v)(v2) + (v)(5) + (3)(v2) + (3)(5)
= v3 + 5v + 3v2 + 15
= v3 + 3v2 + 5v + 15
9-8
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Solutions (continued)
6. (2q2 – 4)(q – 5)
= (2q2)(q) + (2q2)(–5) + (–4)(q) + (–4)(–5)
= 2q3 – 10q2 – 4q + 20
7. (2t – 5)(3t + 4)
= (2t)(3t) + (2t)(4) + (–5)(3t) + (–5)(4)
= 6t2 + 8t – 15t – 20
= 6t2 – 7t – 20
8. (4x – 1)(x2 + 2x + 3) = (4x)(x2) + (4x)(2x) + (4x)(3)
+ (–1)(x2) + (–1)(2x) + (–1)(3)
= 4x3 + 8x2 + 12x – x2 – 2x – 3
= 4x3 + (8 – 1)x2 + (12 – 2)x – 3
= 4x3 + 7x2 + 10x – 3
9-8
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor 6x3 + 3x2 – 4x – 2.
6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1)
= (2x + 1)(3x2 – 2)
Check: 6x3 + 3x2 – 4x – 2
Factor the GCF from each
group of two terms.
Factor out (2x + 1).
(2x + 1)(3x2 – 2)
= 6x3 – 4x + 3x2 – 2
Use FOIL.
= 6x3 + 3x2 – 4x – 2
Write in standard form.
Quick Check
9-8
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor 2t3 + 3t2 + 4t + 6.
2t3 + 3t2 + 4t + 6
t2(2t + 3) + 2(2t + 3)
Factor by grouping.
(2t + 3)(t2 + 2)
Factor again.
Quick Check
9-8
Factoring by Grouping
ALGEBRA 1 LESSON 9-8
Factor each expression.
1. 10p3 – 25p2 + 4p – 10
(5p2 + 2)(2p – 5)
2. 36x4 – 48x3 + 9x2 – 12x
3x(4x2 + 1)(3x – 4)
3. 16a3 – 24a2 + 12a – 18
2(4a2 + 3)(2a – 3)
9-8
Factoring to Solve Quadratic Equations
ALGEBRA 1 Lesson 10-4
Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = 0
x–4=0
or
2x = –3
3
x=–2
Use the Zero-Product Property.
Solve for x.
or
x=4
3
Check: Substitute – 2 for x.
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(– 3 ) + 3](– 3 – 4)
2
2
(2x + 3)(x – 4) = 0
[2(4) + 3](4 – 4)
0
1
(0)(– 5 2 ) = 0
0
(11)(0) = 0
10-4
Quick Check
Factoring to Solve Quadratic Equations
ALGEBRA 1 Lesson 10-4
Solve x2 + 7x – 6x – 42 = 0 by factoring.
x (x+ 7) – 6 (x +7) = 0
(x + 7)(x – 6) = 0
Factor by grouping.
Factor using x2 + 7x – 6x – 42
x+7=0
or
x–6=0
Use the Zero-Product Property.
x = –7
or
x=6
Solve for x.
Quick Check
10-4
Factoring to Solve Quadratic Equations
ALGEBRA 1 Lesson 10-4
Solve 3x2 + 7x – 9x = 21 by factoring.
3x2 + 7x – 9x = 21
Subtract 21 from each side.
3x2 + 7x – 9x – 21= 0
(3x + 7)(x – 3) = 0
3x + 7 = 0
Factor 3x2 + 7x – 9x – 21.
x–3=0
or
3x = –7
7
x=– 3
Use the Zero-Product Property
Solve for x.
or
x=3
Quick Check
10-4