- 105 Views
- Uploaded on
- Presentation posted in: General

Factoring by Grouping

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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 – 42.9r3 + 15r2 + 21r

3.30h3 – 25h2 – 40h4.16m3 – 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

ALGEBRA 1 LESSON 9-8

Solutions

1.6y2 + 12y – 42.9r3 + 15r2 + 21r

6y2 = 2 • 3 • y • y; 9r3 = 3 • 3 • r • r • r;

12y = 2 • 2 • 3 • y; 4 = 2 • 2;15r2 = 3 • 5 • r • r; 21r = 3 • 7 • r;

GCF = 2GCF = 3r

3.30h3 – 25h2 – 40h4.16m3 – 12m2 – 36m

30h3 = 2 • 3 • 5 • h • h • h;16m3 = 2 • 2 • 2 • 2 • m • m • m;

25h2 = 5 • 5 • h • h;12m2 = 2 • 2 • 3 • m • m;

40h = 2 • 2 • 2 • 5 • h;36m = 2 • 2 • 3 • 3 • m;

GCF = 5hGCF = 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

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

Check: 6x3 + 3x2 – 4x – 2 (2x + 1)(3x2 – 2)

= 6x3 – 4x + 3x2 – 2Use FOIL.

= 6x3 + 3x2 – 4x – 2 Write in standard form.

ALGEBRA 1 LESSON 9-8

Factor 6x3 + 3x2 – 4x – 2.

6x3 + 3x2 – 4x – 2 = 3x2(2x + 1) – 2(2x + 1)Factor the GCF from each group of two terms.

= (2x + 1)(3x2 – 2)Factor out (2x + 1).

Quick Check

9-8

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

ALGEBRA 1 LESSON 9-8

Factor each expression.

1.10p3 – 25p2 + 4p – 10

2.36x4 – 48x3 + 9x2 – 12x

3.16a3 – 24a2 + 12a – 18

(5p2 + 2)(2p – 5)

3x(4x2 + 1)(3x – 4)

2(4a2 + 3)(2a – 3)

9-8

2x + 3 = 0 or x – 4 = 0

Use the Zero-Product Property.

2x = –3

Solve for x.

3

2

x = –

or

x = 4

3

2

Check: Substitute – for x.

Substitute 4 for x.

(2x + 3)(x – 4) = 0

(2x + 3)(x – 4) = 0

3

2

3

2

[2(– ) + 3](– – 4) 0

[2(4) + 3](4 – 4) 0

1

2

(11)(0) = 0

(0)(– 5 ) = 0

ALGEBRA 1 Lesson 10-4

Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.

(2x + 3)(x – 4) = 0

Quick Check

10-4

(x + 7)(x – 6) = 0

Factor using x2+ 7x – 6x – 42

x + 7 = 0orx – 6 = 0

Use the Zero-Product Property.

Solve for x.

x = –7orx = 6

ALGEBRA 1 Lesson 10-4

Solve x2 + 7x – 6x– 42 = 0 by factoring.

x (x+ 7) – 6 (x+7) = 0 Factor by grouping.

Quick Check

10-4

3x2 + 7x – 9x = 21

3x2 + 7x – 9x – 21= 0

Subtract 21 from each side.

(3x + 7)(x – 3) = 0

Factor 3x2+ 7x – 9x– 21.

3x + 7 = 0orx – 3 = 0

Use the Zero-Product Property

3x = –7

Solve for x.

7

3

x = – or x = 3

ALGEBRA 1 Lesson 10-4

Solve 3x2 + 7x – 9x = 21 by factoring.

Quick Check

10-4