Chapter Six:   Polynomials and Factoring

Take notes on the following topics.

Chapter Six

Section 6.1

Section 6.2

Section 6.3

Section 6.4

Section 6.5

Section 6.6

Section 6.7

Section 6.8

Section 6.9

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Section 6.4 Factoring  x² + bx + c

(Examples provided by Prentice Hall)

Part I: Constant Term Positive

Factor.

 

Example 1	x2 + 8x + 12 The factorizations of 12 are 1 • 12, 2 • 6, 3 • 4,  –1 • (–12), –2 • (–6), –3 • (–4).  The numbers 2 and 6 are the only pair with the sum 8.  The factorization is (x + 2)(x + 6).
Example 2	x2 – 10x + 16 The factorizations of 16 are 1 • 16, 2 • 8, 4 • 4,  –1 • (–16), –2 • (–8), –4 • (–4).  The factors –2, –8 have sum –10.  The factorization is (x – 2)(x – 8).
Example 3	p2 – 3pq + 2q2 The factorizations of 2 are 1 • 2 and –1 • (–2).  The factors –1 and –2 have sum –3.  The required factorization is (p – 1q)(p – 2q).

Part II: Constant Term Negative

 

Example 1	u2 – 3uv – 10v2 The factorizations of –10 are –1 • 10, –2 • 5, 1 • (–10),  2 • (–5).  The pair 2, –5 has sum –3.  The required factorization is (u + 2v)(u – 5v).
Example 2	x2 + 3x – 4 The factorizations of –4 are 1 • (–4), –1 • 4, 2 • (–2).  The pair –1, 4 has sum 3.  The required factorization is (x – 1)(x + 4).
Example 3	y2 – 12yz – 28z2 The factorizations of –28 are –1 • 28, –2 • 14, –4 • 7, 1 • (–28), 2 • (–14), 4 • (–7).  The pair 2, –14 has sum –12.  The required factorization is (y + 2z)(y – 14z).

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