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

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

4. b. f (2) = 3 (2) ² + 8 (2) – 4 PEMDAS; exponents first f (2) = 3 (4) + 8 (2) – 4 multiply f (2) = – 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. .

9. 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)

10. 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.

11. -6, -3, 2, and 4
P(x) = (x+6)(x+3)(x-2)(x-4)

12. 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) = simplify each parenthesis
c(4)(2)(-1)(-3)(-5) = multiply
-120c = divide both sides by -120
c = 1/30