1. (a) long division (b)synthetic division
Warm up 9/29
Dividing using
(a) long division (b)synthetic division
     x3+3x2-10x-1

2. Agenda: Warmup Notes 6.4 & 6.5 Go over ch 3 test if time
Be seated before the bell rings
Agenda:
Warmup
Notes 6.4 & 6.5
Go over ch 3 test if time
DESK
Warm-up (in your notes)
homework

3. Notebook 6.4 Factor/6.5 Find roots 21) Factor and find roots HW
Table of content
6.4 Factor/6.5 Find roots
Page
19)Add, subtract, multiply polynomials
20) Dividing Polynomials
21) Factor and find roots 1 HW
p.433(17-31 odd)
p.442(15-22)

4. Factor 6.4 Factor/6.5 Find roots 1. f(x)= x2 – 7x 2. f(x)= x2 – 9

5. Factor by grouping 5. x3 – x2 – 25x + 25. (x3 – x2) + (–25x + 25)
Group terms.
x2(x – 1) – 25(x – 1)
Factor GCF from each group.
(x – 1)(x2 – 25)
Factor out the common binomial
(x – 1)(x – 5)(x + 5)
Factor the difference of squares.

6. You try! 6. Factor: x3 – 2x2 – 9x + 18. (x3 – 2x2) + (–9x + 18)

7. (2x3 + x2) + (8x + 4) x2(2x + 1) + 4(2x + 1) (2x + 1)(x2 + 4)
Check It Out!
7. Factor: 2x3 + x2 + 8x + 4.
(2x3 + x2) + (8x + 4)
x2(2x + 1) + 4(2x + 1)
(2x + 1)(x2 + 4)
(2x + 1)(x2 + 4)

8. 8. (x3 + 27)
9. x 6 - 8
(x2)3 - (2)3
(x3 + 33)
( )[( )  + ( )2] x2 2 x2 z 2 2
(x + 3)(x2 – x  )
(x2 - 2 )(x4– 2x + 4)
(x + 3)(x2 – 3x + 9)

9. d3 – 8
(5d)3 – 23
(5d – 2)[(5d)2 + 5d  ]
(5d – 2)(25d2 + 10d + 4)

10. 2x2(x3 – 8) 2x2(x3 – 23) 2x2(x – 2)(x2 + x  2 + 22)

14. Roots of a Polynomial Equation
x-intercept, where graph crosses x-axis
How to “solve” a polynomial equation:
Set the equation equal to 0
Factor the Equation
Set each factor equal to 0 and solve for roots!

15. Find the roots of the equation by factoring.
14. x2 – 4x = –4
x2 – 4x + 4 = 0
(x – 2)(x – 2) = 0
x – 2 = 0 or x – 2 = 0
x = 2 or x = 2

16. Solve the polynomial equation by factoring.
15. 4x6 + 4x5 – 24x4 = 0
4x4(x2 + x – 6) = 0
Factor out the GCF, 4x4.
4x4(x + 3)(x – 2) = 0
Factor the quadratic.
4x4 = 0 or (x + 3) = 0 or (x – 2) = 0
Set each factor equal to 0.
x = 0, x = –3, x = 2
Solve for x.
The roots are 0, –3, and 2.

17. Determine whether the given binomial is a factor of the polynomial P(x).
16. (x + 1); (x2 – 3x + 1)
17. (x + 2); (3x4 + 6x3 – 5x – 10)
Find P(–1) by synthetic substitution.
Find P(–2) by synthetic substitution. –1
1 –3 1 –2
–5 –10 –1 4 –6 10 1 –4 5 3 –5
P(–1) = 5
P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.
P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1.

18. 18. x3 + 3x2 – 4x – 12 = 0 Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12
Steps:
Identify possible roots
Test to find one that makes the equation true
That is a factor, divide to see what’s left
Factor more if necessary or use quadratic formula if possible
18.
x3 + 3x2 – 4x – 12 = 0
Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12

19. 19. Identify all the real roots of 2x3 – 3x2 –10x – 4 = 0. Steps:
Identify possible roots
Test to find one that makes the equation true
That is a factor, divide to see what’s left
Factor more if necessary or use quadratic formula if possible

20. 20. Identify all the real roots of 2x3 – 42x+40 = 0. Steps:
Identify possible roots
Test to find one that makes the equation true
That is a factor, divide to see what’s left
Factor more if necessary or use quadratic formula if possible

21. 19. Identify all the real roots of 4x5 – 8x3+4x = 0. Steps:
Identify possible roots
Test to find one that makes the equation true
That is a factor, divide to see what’s left
Factor more if necessary or use quadratic formula if possible

22. Divide the polynomial by 3, then find P(2) by synthetic substitution.
Check It Out! Example 1
Determine whether the given binomial is a factor of the polynomial P(x).
a. (x + 2); (4x2 – 2x + 5)
b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30)
Find P(–2) by synthetic substitution.
Divide the polynomial by 3, then find P(2) by synthetic substitution. –2
4 –2 5 –8 20 2
1 – –10 4
–10 25 2 4 10
P(–2) = 25 1 2 5
P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.
P(2) = 0, so (3x – 6) is a factor of P(x) = 3x4 – 6x3 + 6x2 + 3x – 30.

25. Solve the polynomial equation by factoring.
2x6 – 10x5 – 12x4 = 0
2x4(x2 – 5x – 6) = 0
Factor out the GCF, 2x4.
2x4(x – 6)(x + 1) = 0
Factor the quadratic.
2x4 = 0 or (x – 6) = 0 or (x + 1) = 0
Set each factor equal to 0.
x = 0, x = 6, x = –1
Solve for x.
The roots are 0, 6, and –1.

26. Solve the polynomial equation by factoring.
x = 26x2
x4 – 26 x = 0
Set the equation equal to 0.
Factor the trinomial in quadratic form.
(x2 – 25)(x2 – 1) = 0
(x – 5)(x + 5)(x – 1)(x + 1)
Factor the difference of two squares.
x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0
x = 5, x = –5, x = 1 or x = –1
Solve for x.
The roots are 5, –5, 1, and –1.