1. Warm Up #1
1. Use synthetic substitution to evaluate f (x) = x3 + x2 – 3x – 10 when x = 2. – 2 6 6 1 3 3 -4
ANSWER –4

2. Check HW 5.4 multiples of 3

4. EXAMPLE 1
Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.

5. ) EXAMPLE 1 Use polynomial long division 3x2 + 4x – 3 x2 – 3x + 5
quotient
x2 – 3x + 5
3x4 – 5x3 + 0x2 + 4x – 6 )
-(3x4 – 9x3 + 15x2)
4x3 – 15x x
-(4x3 – 12x2 + 20x)
–3x2 – 16x – 6
-(–3x2 + 9x – 15)
–25x + 9
remainder
3x4 – 5x3 + 4x – 6
x2 – 3x + 5
= 3x2 + 4x – 3 +
–25x + 9
ANSWER

6. ) EXAMPLE 2 Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2. x2
+ 7x
+ 7
quotient
x – 2
x3 + 5x2 – 7x )
-(x3 – 2x2)
7x2 – 7x
-(7x2 – 14x)
7x + 2
-(7x – 14) 16
remainder
ANSWER
x3 + 5x2 – 7x +2
x – 2
= x2 + 7x + 7 + 16

7. GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
(2x4 + x3 + x – 1) (x2 + 2x – 1)
(2x2 – 3x + 8) +
–18x + 7
x2 + 2x – 1
ANSWER
(x3 – x2 + 4x – 10)  (x + 2)
(x2 – 3x + 10) +
–30
x + 2
ANSWER

8. GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
(x3 – x2 + 4x – 10)  (x + 2)
(x2 – 3x + 10) +
–30
x + 2
ANSWER

9. EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
– – -6 15
-21 2 -5 7
-16

10. EXAMPLE 4
Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that
x + 2 is a factor.
SOLUTION – -6 20 16 3
-10 -8

11. Use the result to write f (x) as a product of two
EXAMPLE 4
Factor a polynomial
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16
Write original polynomial.
= (x + 2)(3x2 – 10x – 8)
Write as a product of two
factors.
= (x + 2)(3x + 2)(x – 4)
Factor trinomial.

12. GUIDED PRACTICE
for Examples 3 and 4
Divide using synthetic division.
3. (x3 + 4x2 – x – 1)  (x + 3) – -3 -3 12 1 1 -4 11
x2 + x – 4 + 11
x + 3
ANSWER

13. (x – 4)(x –3)(x + 1) GUIDED PRACTICE for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a factor.
f (x) = x3 – 6x2 + 5x + 12
(x – 4)(x –3)(x + 1) 4 -8
-12 1 -2 -3

14. Zeros are -2, 3, -3 GUIDED PRACTICE for Examples 5 and 6
Find the other zeros of f given that f (–2) = 0.
f (x) = x3 + 2x2 – 9x – 18 – -2 18
Zeros are -2, 3, -3 1 -9

15. EXAMPLE 5
Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division.
–2 –
–60 –

16. EXAMPLE 5
Standardized Test Practice
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
= (x – 3)(x2 + x – 20)
= (x – 3)(x + 5)(x – 4)
The zeros are 3, –5, and 4.
The correct answer is A.
ANSWER

17. Class/Homework Assignment WS 5.5 (1-24 mult. of 3)