9.8

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  • End of day 1
  • 1. (x-3)(x+3) 2. 4(y+4)(y-4) 3. (h+2)^2 4. 2(y-5)^2
  • 1. -3 2. 7 3. 9, -9
  • 9.8

    1. 1. Difference of Two Squares (Factor) a - b = (a + b)(a - b) 2 2 y -16 = y - 4 = (y + 4)(y - 4) 2 2 2
    2. 2. Example 1 Factor the difference of two squares Factor the polynomial. a. 25m2 – 36 = ( 5m )2 – 62 Write as a2 – b2. = (5m + 6 ) (5m – 6 ) Difference of two squares pattern b. x2 – 49y2 = x2 – ( 7y)2 = (x + 7y ) (x – 7y) c. 8 – 18n2 = 2( 4 – 9n2 ) = 2[22 – ( 3n)2] Write as a2 – b2. Difference of two squares pattern Factor out common factor. Write 4 – 9n2 as a2 – b2.
    3. 3. Example 1 Factor the difference of two squares = 2(2 + 3n ) (2 – 3n) d. – 9 + 4x2 = 4x2 – 9 Difference of two squares pattern Rewrite as difference. = ( 2x )2 – 32 Write as a2 – b2. = (2x + 3 ) (2x – 3 ) Difference of two squares pattern
    4. 4. Perfect Square Trinomial x 2 + 6x + 9 a + 2ab + b 2 x 2 -10x + 25 2 a - 2ab + b 2 2
    5. 5. Perfect Square Trinomial (Factor) a2 + 2ab + b2 = (a + b)2 Algebra Example x + 6x + 9 = x + 2(x ×3) + 3 = (x + 3) Algebra (a - b)2 a - 2ab + b = 2 2 2 2 2 2 (x - 5)2 Example x -10x + 25 = x - 2(x × 5)+ 5 = 2 2 2
    6. 6. Example 2 Factor perfect square trinomials Factor the polynomial. a. n2 – 12n + 36 = n2 – 2( n • 6) + 62 = (n – 6 )2 Write as a2 – 2ab + b2. Perfect square trinomial pattern b. 9x2 – 12x + 4 = ( 3x)2 – 2( 3x • 2) + 22 Write as a2 – 2ab + b2. = (3x – 2 )2 Perfect square trinomial pattern c. 4s2 + 4st + t2 = ( 2s )2 + 2( 2s • t ) + t2 Write as a2 + 2ab + b2. = (2s + t )2 Perfect square trinomial pattern
    7. 7. Example 3 Multiple Choice Practice What is the factored form of – 3x2 + 36xy – 108y2 ? ( – 3x – 6y )2 – 3( x + 6y )2 ( 3x – 6y )2 – 3( x – 6y )2 SOLUTION –3x2 + 36xy – 108y2 = –3( x2 – 12xy + 36y2 ) Factor out – 3 . 2 = –3[x2 – 2 (x • 6y ) + ( 6y) ] Write x2 – 12xy + 36y2 as a2 – 2ab + b2.
    8. 8. Example 3 Multiple Choice Practice = – 3( x – 6y )2 ANSWER Perfect square trinomial pattern The correct answer is D.
    9. 9. Example 4 Solve a polynomial equation Solve the equation x2 x2 1 x + + = 0. 3 9 = 0 Write original equation. 6x + 1 = 0 Multiply each side by 9. 2 1 x + + 3 9x2 + 2 9 ( 3x )2 + 2 ( 3x • 1 ) + ( 1 )2 = 0 Write left side as a2 + 2ab + b2. ( 3x + 1 )2 = 0 Perfect square trinomial pattern 3x + 1 = 0 x = – Zero-product property 1 3 Solve for x.
    10. 10. Example 4 ANSWER Solve a polynomial equation The solution of the equation is – 1 3 .
    11. 11. Example 5 Solve a vertical motion problem 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. Because the sponge was dropped, it has no initial vertical velocity. To determine when the sponge lands on the ground, find the value of t for which the height is 0.
    12. 12. Example 5 Solve a vertical motion problem h = – 16t2 + vt + s Vertical motion model 0 = – 16t2 + ( 0 ) t + 64 Substitute 0 for h, 0 for v, and 64 for s. 0 = – 16 ( t2 – 4 ) Factor out – 16 . 0 = – 16 ( t + 2 ) ( t – 2 ) Difference of two squares pattern t + 2 = 0 or t = –2 or t – 2 = 0 t = 2 Zero-product property Solve for t. Disregard the negative solution of the equation.
    13. 13. Example 5 Solve a vertical motion problem ANSWER The sponge lands on the ground 2 seconds after it is dropped.
    14. 14. 9.8 Warm-Up (Day 1) Factor the polynomial. 1. x2 - 9 2. 4y 2 - 64 3. h + 4h + 4 4. 2y2 - 20y + 50 2
    15. 15. 9.8 Warm-Up (Day 2) Solve the equation. 1. a2 + 6a + 9 = 0 2. w -14w + 49 = 0 3. n -81= 0 2 2

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