Eureka Math Algebra 2 Module 1 Lesson 19

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Presentation transcript:

Eureka Math Algebra 2 Module 1 Lesson 19 Remainder Theorem Eureka Math Algebra 2 Module 1 Lesson 19

Objectives Students know and apply the remainder theorem and understand the role zeros play in the theorem. Standard A.APR.B.2

b. f (2) = 3 (2) ² + 8 (2) – 4 PEMDAS; exponents first f (2) = 3 (4) + 8 (2) – 4 multiply f (2) = 12 + 16 – 4 add/subtract f (2) = 24 What do you notice about the remainder and the value of the function? They are both the same with the value of 24. .

x ²-5x+6 (x-2)(x-3) Plug in -1 for x, simplify and collect like terms Solve for k by subtracting 4 on both sides x ²-5x+6 (x-2)(x-3)

X+3 is a factor of P since P(-3) = 0. Since P(1) = 84, 1 is not a zero. X+3 is a factor of P since P(-3) = 0.

-6, -3, 2, and 4 P(x) = (x+6)(x+3)(x-2)(x-4)

f(x) = c (x+4) (x+2) (x-1) (x-3) (x-5) Since the y-intercept is -4, that means f(0) = -4 f(0) = c (0+4)(0+2)(0-1)(0-3)(0-5) = -4 simplify each parenthesis c(4)(2)(-1)(-3)(-5) = -4 multiply -120c = -4 divide both sides by -120 c = 1/30