Section 5.5 – The Real Zeros of a Rational FunctionRemainder Theorem

Example:

If f(x) is a polynomial function and is divided by x – c, then the remainder is f(c).

𝑓 (𝑥 )=𝑥2−2 𝑥−15

𝑓 (4 )=42−2 (4 )−15The remainder after dividing f(x) by (x – 4) would be -7.

𝐷𝑖𝑣𝑖𝑑𝑒𝑏𝑦 h𝑡 𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 :𝑥−4𝑓 (4 )=−7

𝑜𝑟 𝑥=4

1

15214 428−7

Section 5.5 – The Real Zeros of a Rational FunctionFactor 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 )=𝑥2−2 𝑥−15

𝑓 (−3 )=(−3)2−2 (−3 )−15The remainder after dividing f(x) by (x + 3) would be 0.

𝐷𝑖𝑣𝑖𝑑𝑒𝑏𝑦 h𝑡 𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 :𝑥+3𝑓 (−3 )=0𝑜𝑟 𝑥=−3

1

15213 −3−5

150

Section 5.5 – The Real Zeros of a Rational FunctionRational Zeros Theorem (for functions of degree 1 or higher)

(2) Each coefficient is an integer.(1)

If (in lowest terms) is a rational zero of the function, then p is a factor of and q is a factor of .

Given:

Theorem: A polynomial function of odd degree with real coefficients has at least one real zero.

Section 5.5 – The Real Zeros of a Rational Function

Example: Find the solution(s) of the equation.

𝑓 (𝑥 )=𝑥3−2 𝑥2−5 𝑥+6𝑝 :±1 , ±2 , ±3 , ±6𝑞 :±1

Possible solutions:

Try:

Rational Zeros Theorem

𝑝𝑞:±11,±21, ±31, ±61

Section 5.5 – The Real Zeros of a Rational Function𝑓 (𝑥 )=𝑥3−2 𝑥2−5 𝑥+6Long Division

1

𝑥2

6521 23 xxxx

Synthetic Division

𝑥3 −𝑥2

−𝑥2 −5 𝑥

−𝑥

−𝑥2 +𝑥−6 𝑥+6

−6

−6 𝑥+60

(𝑥−1 ) (𝑥2−𝑥−6 )

65211 1−1

−1−6

−60

(𝑥−1 ) (𝑥2−𝑥−6 )

Section 5.5 – The Real Zeros of a Rational Function𝑓 (𝑥 )=𝑥3−2 𝑥2−5 𝑥+6

(𝑥−1 ) (𝑥2−𝑥−6 )=0(𝑥−1 )(𝑥+2 )(𝑥−3 )¿0

(𝑥−1 )=0 (𝑥+2 )=0 (𝑥−3 )=0

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 :𝑥=−2 ,1 ,3

Section 5.5 – The Real Zeros of a Rational FunctionExample: Find the solution(s) of the equation.

𝑓 (𝑥 )=4 𝑥4+7 𝑥2−2 𝑝 :±1 , ±2𝑞 :±1 ,±2 , ±4Possible solutions :

Try:

4

207041 444111111119

Try:

4

207042 88162346469290

Try:

4

2070421 22184420

(𝑥−12)(4 𝑥3+2 𝑥2+8 𝑥+4)

Section 5.5 – The Real Zeros of a Rational Function

(𝑥−12)(4 𝑥3+2 𝑥2+8 𝑥+4)

(𝑥−12)(2)(2𝑥3+𝑥2+4 𝑥+2)

(𝑥−12)(2)(𝑥2 (2𝑥+1 )+2 (2 𝑥+1 ))

(𝑥−12)(2)(𝑥2+2) (2𝑥+1 )

(𝑥− 12 )=02𝑥2+2=02𝑥+1=0

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 : 𝑥=−12,12

Section 5.5 – The Real Zeros of a Rational FunctionIntermediate 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.

(𝑎 , 𝑓 (𝑎 ))

(𝑏 , 𝑓 (𝑏)) (𝑎 , 𝑓 (𝑎 ))

(𝑏 , 𝑓 (𝑏))𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜

𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜

Section 5.5 – The Real Zeros of a Rational Function

𝑓 (𝑥 )=2𝑥3−3 𝑥2−2

𝑓 (0 )=¿

Intermediate Value TheoremDo the following polynomial functions have at least one real zero in the given interval?

−2 𝑓 (2 )=¿2𝑦𝑒𝑠

[0 ,2]𝑓 (𝑥 )=2𝑥3−3 𝑥2−2

𝑓 (3 )=¿25 𝑓 (6 )=¿322𝑛𝑜𝑡 h𝑒𝑛𝑜𝑢𝑔 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛

[3 ,6]

𝑓 (𝑥 )=𝑥4−2𝑥2−3 𝑥−3

𝑓 (−5 )=¿587 𝑓 (−2 )=¿11𝑛𝑜𝑡 h𝑒𝑛𝑜𝑢𝑔 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛

[−5 ,−2]𝑓 (𝑥 )=𝑥4−2𝑥2−3 𝑥−3

𝑓 (−1 )=¿−1𝑓 (3 )=¿51𝑦𝑒𝑠

[−1 ,3]