1. Section 5.5 – The Real Zeros of a Rational Function
Remainder Theorem
If f(x) is a polynomial function and is divided by x – c, then the remainder is f(c).
Example:
𝑓 𝑥 = 𝑥 2 −2𝑥−15
𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑥−4
𝑜𝑟 𝑥=4
𝑓 4 = 4 2 −2 4 −15
𝑓 4 =−7
The remainder after dividing f(x) by (x – 4) would be -7. 4 8 1 2 −7

2. Section 5.5 – The Real Zeros of a Rational Function
Factor Theorem
If f(x) is a polynomial function, then x – c is a factor of f(x) if and only if f(c) = 0.
Example:
𝑓 𝑥 = 𝑥 2 −2𝑥−15
𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑥+3
𝑜𝑟 𝑥=−3
𝑓 −3 = (−3) 2 −2 −3 −15
𝑓 −3 =0
The remainder after dividing f(x) by (x + 3) would be 0. −3 15 1 −5

3. Section 5.5 – The Real Zeros of a Rational Function
Rational Zeros Theorem (for functions of degree 1 or higher)
Given:
(1) 𝑓 𝑥 = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + ⋯ 𝑎 1 𝑥+ 𝑎 0
(2) Each coefficient is an integer.
If 𝑝 𝑞 (in lowest terms) is a rational zero of the function, then p is a factor of 𝑎 0 and q is a factor of 𝑎 𝑛 .
Theorem: A polynomial function of odd degree with real coefficients has at least one real zero.

4. Section 5.5 – The Real Zeros of a Rational Function
Rational Zeros Theorem
Example: Find the solution(s) of the equation.
𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6
𝑝: ±1, ±2, ±3, ± 𝑞: ±1
𝑝 𝑞 : ± 1 1 , ± 2 1 , ± 3 1 , ± 6 1
Possible solutions: 𝑥=±1, ±2, ±3, ±6
Try: 𝑥= 𝑜𝑟 𝑥−1=0

5. Section 5.5 – The Real Zeros of a Rational Function
𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6
Long Division
Synthetic Division
𝑥 2 −𝑥 −6
𝑥 3
−𝑥 2 1 −1 −6
−𝑥 2
−5𝑥 1 −1 −6
−𝑥 2 +𝑥
−6𝑥 +6
𝑥−1 𝑥 2 −𝑥−6
−6𝑥 +6
𝑥−1 𝑥 2 −𝑥−6

6. Section 5.5 – The Real Zeros of a Rational Function
𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6
𝑥−1 𝑥 2 −𝑥−6 =0
𝑥−1
𝑥+2
𝑥−3 =0
𝑥−1 =0
𝑥+2 =0
𝑥−3 =0
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠:
𝑥=−2, 1, 3

7. Section 5.5 – The Real Zeros of a Rational Function
Example: Find the solution(s) of the equation.
𝑓 𝑥 = 4𝑥 4 +7 𝑥 2 −2
𝑝: ±1, ± 𝑞: ±1, ±2, ±4
Possible solutions ( 𝑝 𝑞 ): 𝑥=±1, ± 1 2 , ± 1 4 , ±2
Try: 𝑥=1
Try: 𝑥=2 4 4 11 11 8 16 46 92 4 4 11 11 9 4 8 23 46 90
Try: 𝑥= 1 2
(𝑥− 1 2 )(4 𝑥 3 +2 𝑥 2 +8𝑥+4) 2 1 4 2 4 2 8 4

8. Section 5.5 – The Real Zeros of a Rational Function
(𝑥− 1 2 )(4 𝑥 3 +2 𝑥 2 +8𝑥+4)
(𝑥− 1 2 )(2)(2 𝑥 3 + 𝑥 2 +4𝑥+2)
(𝑥− 1 2 )(2)( 𝑥 2 2𝑥 𝑥+1 )
(𝑥− 1 2 )(2)( 𝑥 2 +2) 2𝑥+1
𝑥− 1 2 = 𝑥 2 +2= 𝑥+1=0
𝑥=− 1 2 , 1 2
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠:

9. Section 5.5 – The Real Zeros of a Rational Function
Intermediate Value Theorem
In a polynomial function, if a < b and f(a) and f(b) are of opposite signs, then there is at least one real zero between a and b.
(𝑏, 𝑓 𝑏 )
(𝑎, 𝑓 𝑎 )
𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜
𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜
(𝑏, 𝑓 𝑏 )
(𝑎, 𝑓 𝑎 )

10. Section 5.5 – The Real Zeros of a Rational Function
Intermediate Value Theorem
Do the following polynomial functions have at least one real zero in the given interval?
𝑓 𝑥 =2 𝑥 3 −3 𝑥 2 −2
𝑓 𝑥 =2 𝑥 3 −3 𝑥 2 −2
[0, 2]
[3, 6]
𝑓 0 = −2
𝑓 2 = 2
𝑓 3 = 25
𝑓 6 =
322
𝑦𝑒𝑠
𝑛𝑜𝑡 𝑒𝑛𝑜𝑢𝑔ℎ 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
𝑓 𝑥 = 𝑥 4 −2 𝑥 2 −3𝑥−3
𝑓 𝑥 = 𝑥 4 −2 𝑥 2 −3𝑥−3
[−5, −2]
[−1, 3]
𝑓 −5 =
587
𝑓 −2 = 11
𝑓 −1 = −1
𝑓 3 = 51
𝑛𝑜𝑡 𝑒𝑛𝑜𝑢𝑔ℎ 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
𝑦𝑒𝑠