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,

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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.

Homework Text p. 578, #4-48 multiples of 4  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 5 4

Objective  SWBAT factor trinomials of the form x² + bx + c

Section 9.4 “Solve Polynomial Equations in Factored Form” 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 is ZERO and the other side is the product of polynomial factors. (x – 4)(x + 2) = 0 Zero-Product Property The solutions of such an equation are called ROOTS. x – 4 = 0 x + 2 = 0 x = 4 x = -2

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

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

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

Factoring polynomials Factors of 18Sum of factors 18, = 19 2, = 11 3, = 9 x² + 11x + 18 (x + 2)(x + 9) Find two factors of 18 whose sum is 11.

Factoring polynomials…Try it out… Solve the equation Factors of 10Sum of factors 10, = 11 2, = 7 x² + 7x + 10 (x + 2)(x + 5)=0 Find two factors of 10 whose sum is 7. x + 2 = 0 x = -2 x + 5 = 0 x = -5

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

Factoring polynomials ‘-’ factors of 8 Sum of factors -8, (-1)= -9 -2, (-4) = -6 n² – 6n + 8 (n – 2)(n – 4) Find two ‘negative’ factors of 8 whose sum is -6.

Factoring polynomials Factors of -15Sum of factors -15, = , (-1) = 14 -3, = 2 3, (-5) = -2 y² + 2y – 15 (y – 3)(y + 5) Find two factors of -15 with different signs whose sum is positive 2.

Factoring the polynomial. Then solve the equation. Factors of -30Sum of factors -30, = , (-1) = 29 -2, = 13 2, (-15) = , = 7 3, (-10) = -7 -5, = 1 5, (-6) = -1 x² - 7x - 30 (x + 3)(x – 10) Find two factors of -30 with different signs whose sum is -7. x = -3, 10

Challenge… Factoring polynomials with two variables Factors of 14Sum of factors 14, = 15 2, = 9 x² + 9xy + 14y² (x + 2y)(x + 7y) Find two factors of 14 whose sum is positive 9.

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

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