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Do Now 3/3/10

Do Now 3/3/10. Take out HW from last night. Text p. 578, #4-48 multiples of 4 Copy HW in your planner. Text p. 586, #4-52 multiples of 4 In your notebook, factor out the GCF monomial factor. 4) -9, 1 8) 1/8, -18 12) -1/3, -6 16) z = 15, -21 20) d (5d + 2) 24) 4a(3a + 2)

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Do Now 3/3/10

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  1. Do Now 3/3/10 • Take out HW from last night. • Text p. 578, #4-48 multiples of 4 • Copy HW in your planner. • Text p. 586, #4-52 multiples of 4 • In your notebook, factor out the GCF monomial factor.

  2. 4) -9, 1 8) 1/8, -18 12) -1/3, -6 16) z = 15, -21 20) d (5d + 2) 24) 4a(3a + 2) 28) 0, 1 32) 0, -3/4 36) 0, -1/3 40) 4xy(5xy – 1) 44) -2g(g³ - 7g – 3) 48) 0, 9 Homework Text p. 578, #4-48 multiples of 4 5 4

  3. Objective • SWBAT factor trinomials of the form x² + bx + c

  4. Section 9.4 “Solve Polynomial Equations in Factored Form” Zero-Product Property If ab = 0, then a = 0 or b = 0. The zero-product property is used to solve an equation when one side of the equation isZEROand the other side is theproduct of polynomial factors. (x – 4)(x + 2) = 0 The solutions of such an equation are called ROOTS. x + 2 = 0 x – 4 = 0 x = -2 x = 4

  5. Solve the equations (x – 5)(x + 1) = 0 x + 1 = 0 x – 5 = 0 x = -1 x = 5 (2x – 3)(4x + 1) = 0 4x + 1 = 0 2x – 3 = 0 x = -1/4 x = 3/2

  6. Solve Equations By Factoring 2x² + 8x = 0 Factor left side of equation 2x(x + 4) = 0 Zero product property x + 4 = 0 2x = 0 x = -4 x = 0 The solutions of the equation are 0 and -4.

  7. Section 9.5 “Factor x² + bx + c” Factoring x² + bx + c x² + bx + c = (x + p)(x + q) provided p + q = b and pq = c x² + 5x + 6 = (x + 3)(x + 2) Remember FOIL

  8. Factoring polynomials x² + 11x + 18 Find two factors of 18 whose sum is 11. (x + 2)(x + 9)

  9. Factoring polynomials…Try it out…Solve the equation x² + 7x + 10 Find two factors of 10 whose sum is 7. (x + 2)(x + 5)=0 x + 2 = 0 x = -2 x + 5 = 0 x = -5

  10. Factoring polynomials When factoring a trinomial, first consider the signs of p and q. x² + bx + c b is positive; c is positive b is negative; c is negative (x + 2)(x – 3) x² – bx – c b is positive; c is negative (x – 2)(x + 3) x² + bx – c (x – 2)(x – 3) x² – bx + c b is negative; c is positive

  11. Factoring polynomials n² – 6n + 8 Find two ‘negative’ factors of 8 whose sum is -6. (n – 2)(n – 4)

  12. Factoring polynomials y² + 2y – 15 Find two factors of -15 with different signs whose sum is positive 2. (y – 3)(y + 5)

  13. Factoring the polynomial. Then solve the equation. x² - 7x - 30 Find two factors of -30 with different signs whose sum is -7. (x + 3)(x – 10) x = -3, 10

  14. Challenge… Factoring polynomials with two variables x² + 9xy + 14y² Find two factors of 14 whose sum is positive 9. (x + 2y)(x + 7y)

  15. Problem Solving Area = length x width Find the dimensions of the rectangle. The area is 100 square inches. (w – 15) w Substitute the solution to see the dimensions of the rectangle. w = 20 w = -5 w = 20 (20-15=5) Can’t have negative width 20 Area = 20 x 5

  16. Homework Text p. 586, #4-52 multiples of 4

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