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Consider the function f(x) = 2x^2 + 5x^3 - 21x^2  - 36x

A) Explaing why the rational zero theorem cannot be directly applied to this function.

B) Factor out the common monomial factor of f.

C) Apply the rational zero theorem to find all the real zeros of f.

D) Find all the real zeros of f(x) = 3x^5 - x^4 - 6x^3 + 2x^2

 Jan 29, 2019
 #1
avatar+99523 
+2

Consider the function f(x) = 2x^2 + 5x^3 - 21x^2  - 36x

A) Explaing why the rational zero theorem cannot be directly applied to this function.

 

The last term must be a constant to apply the theorem

 

 

B) Factor out the common monomial factor of f.

 

 x (2x + 5x^2 - 21x - 36) = 0

 

x ( 5x^2 - 19x - 36) = 0

 

The possible rational zeoroes  are all the possible divisors of 36   ovet all the possibe divisors of 5

 

However....the only rational zero here is x =0

The other two solutions involve non-rational  (but real) answers

 

 

cool cool cool

 Jan 29, 2019
 #2
avatar+99523 
+2

D) Find all the real zeros of f(x) = 3x^5 - x^4 - 6x^3 + 2x^2

 

Factor out x^2

 

x^2 (3x^3 - x^2 - 6x + 2)

 

One zero = 0

 

Factors of 2 =  1 , 2

Factors of 3 = 1 . 3

 

So...The other possible rational roots are   ± [ 1,  2, 1/3 , 2/3 ]

 

A little trial and error shows that the only other rational root = 1/3

 

The other two roots are non-rational reals

 

 

cool cool cool

 Jan 29, 2019

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