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Factor Special Products

8.7. Factor Special Products. Difference of Two Squares. Algebra a 2 –b 2 = (a + b)(a -b) Example 4x 2 – 9 = (2x) 2 +3 2 = (2x + 3)(2x -3). EXAMPLE 1. Factor the difference of two squares. Factor the polynomial. a . y 2 – 16 = y 2 – 4 2. Write as a 2 – b 2.

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Factor Special Products

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  1. 8.7 Factor Special Products

  2. Difference of Two Squares Algebra a2 –b2 = (a + b)(a -b) Example 4x2 – 9 = (2x)2 +32 = (2x + 3)(2x -3)

  3. EXAMPLE 1 Factor the difference of two squares Factor the polynomial. a. y2 – 16 = y2 – 42 Write as a2 – b2. = (y + 4)(y – 4) Difference of two squares pattern b. 25m2 – 36 = (5m)2 – 62 Write as a2 – b2. = (5m + 6)(5m – 6) Difference of two squares pattern c. x2 – 49y2 = x2 – (7y)2 Write as a2 – b2. = (x + 7y)(x – 7y) Difference of two squares pattern

  4. EXAMPLE 2 Factor the difference of two squares Factor the polynomial 8 – 18n2. 8 – 18n2 = 2(4 – 9n2) Factor out common factor. = 2[22 – (3n) 2] Write 4 – 9n2asa2–b2. = 2(2 + 3n)(2 – 3n) Difference of two squares pattern

  5. for Examples 1 and 2 GUIDED PRACTICE Factor the polynomial. 1. 4y2 – 64 = (2y + 8)(2y – 8) 2. 4m2 – n2 = (2m + n)(2m– n) 3. 18x2 – 8 = 2(9x2 - 4) = 2(3x + 2)(3x – 2)

  6. Perfect Square Trinomials Algebra Example a2 +2ab+ b2 = (a + b)2x2 + 6x + 9 = x2+ 2(x)(3) + 32 = (x+3)2 a2 -2ab+ b2 = (a - b)2x2 - 10x + 25 = x2 - 2(x)(5) + 52 = (x-5)2

  7. a. n2 – 12n + 36 = n2 – 2(n 6) + 62 9x2 – 12x + 4 b. c. 4s2 + 4st + t2 = (2s)2 + 2(2s t) + t2 Factor perfect square trinomials EXAMPLE 3 Factor the polynomial. Write asa2 – 2ab + b2. = (n – 6)2 Perfect square trinomial pattern = (3x)2 – 2(3x 2) + 22 Write asa2 – 2ab + b2. =(3x – 2)2 Perfect square trinomial pattern Write asa2 + 2ab + b2. = (2s + t)2 Perfect square trinomial pattern

  8. =–3[y2 – 2(y 6) + 62] Factor a perfect square trinomial EXAMPLE 4 Factor the polynomial–3y2 + 36y – 108. Factor out–3. –3y2 + 36y – 108 = –3(y2 – 12y + 36) Writeasa2 – 2ab + b2. = –3(y – 6)2 Perfect square trinomial

  9. for Examples 3 and 4 GUIDED PRACTICE Factor the polynomial. 1.h2 + 4h + 4 = (h + 2)2 2. 2y2 – 20y + 50 = 2(y –5)2 3. 3x2 + 6xy + 3y2 = 3(x + y)2 4. 4y2 - 16y + 16 = (4y - 4)2

  10. ANSWER 1 1 The solution of the equation is–. (3x)2 + 2(3x 1) + (1)2 = 0 3 3 x =– Solve a polynomial equation EXAMPLE 5 Solve the equation 9x2 + 6x + 1=0. 9x2+6x+1=0 Write left side asa2 + 2ab + b2. (3x + 1)2 = 0 Zero-product property Solve forx.

  11. Solve a vertical motion problem EXAMPLE 6 FALLING OBJECT A window washer drops a wet sponge from a height of 64 feet. After how many seconds does the sponge land on the ground? SOLUTION Use the vertical motion model to write an equation for the height h(in feet) of the sponge as a function of the time t(in seconds) after it is dropped.

  12. t = 2 or t = –2 or t – 2 = 0 t + 2 = 0 The sponge lands on the ground 2 seconds after it is dropped. ANSWER Solve a vertical motion problem EXAMPLE 6 The sponge was dropped, so it has no initial vertical velocity. Find the value of t for which the height is 0. h = –16t2 + vt + s Vertical motion model 0 = –16t2 + (0)t + 64 Substitute 0 for h,0 for v, and 64 for s. Factor out–16. 0 = –16(t2 – 4) 0 = –16(t – 2)(t +2) Difference of two squares pattern Zero-product property Solve fort. Disregard the negative solution of the equation.

  13. n = – 9 or n = 9 for Examples 5 and 6 GUIDED PRACTICE Solve the equation a = –3 5. a2+6a+ 9 = 0 6. w2–14w+49 = 0 w = 7 7. n2–81=0

  14. The sponge lands on the ground 1 second after it is dropped. ANSWER for Examples 5 and 6 GUIDED PRACTICE 8. WHAT IF? In Example 6, suppose the sponge is dropped from a height of 16 feet. After how many seconds does it land on the ground?

  15. (2m – n)(2m + n) ANSWER ANSWER (x + 3)2 (2y – 4)2 ANSWER Daily Homework Quiz Factor the trinomial. 1. 4m2 – n2 2.x2 + 6x + 9 3. 4y2 – 16y+16

  16. x2 + x + = 0 – ANSWER 1 1 4 2 An apple falls from a branch 9 feet above the ground. After how many seconds does the apple hit the ground? 5. ANSWER 0.75 sec Daily Homework Quiz 4.Solve the equation

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