1. Warm Up

2. Objective: Students will divide polynomials

3. Use long division to divide 125 by 3
2304 ÷ 7

4. Divide x2 + 2x – 30 by x – 5. Dividing Polynomials Additional Examples
LESSON 6-3
Additional Examples
Divide x2 + 2x – 30 by x – 5. 

6. Determine whether x + 2 is a factor of each polynomial.
Dividing Polynomials
LESSON 6-3
Additional Examples
Determine whether x + 2 is a factor of each polynomial.
a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6
x2 + 5x – 15
x + 2 x3 + 7x2 – 5x – 6
x3 + 2x2
5x2 – 5x
5x2 + 10x
–15x – 6
–15x – 30 24
x + 8
x + 2 x2 + 10x + 16
x2 + 2x
8x + 16
Since the remainder is zero,
x + 2 is a factor of x2 + 10x + 16.
Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6. /

8. Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.
Dividing Polynomials
LESSON 6-3
Additional Examples
Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.
Step 1: Reverse the sign of the constant term in the divisor.
Write the coefficients of the polynomial in standard form. –
Write x x3 – 6x2 + 4x – 1 3 as
5 – – 1
Step 2: Bring down the coefficient.
Bring down the 5.
3 5 –6 4 –1 This begins the quotient. 5

10. Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.
Dividing Polynomials
LESSON 6-3
Additional Examples
Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.
By the Remainder Theorem, P(3) equals the remainder when
P(x) is divided by x – 3.
3 1 – –9
The remainder is 21, so P(3) = 21.

11. Find P(4) is P(x)= 2x³ - 4x² + x – 5
Find P(-3), P(x)= x³ -5x² + 8

12. Objective: Students will be able to solve polynomials equations

15. Factor
2: 8x³ -1
Solve
3: 27x³ + 1=0
4: x³ + 8 = 0