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  #1  
Old January 5th, 2009, 07:18 AM
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Default factor theorem

Given that (x-3) and (2x+1) are factors of
f(x)=ax^4+bx^3+13x^2=30x+9, find the values of a and b . With these values of a and b , show that f(x)\geq0 for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

4x^4-20x^3+13x^2+30x+9
(x-3)(2x+1)(2x^2-5x-3)

I am wondering how can i show that f(x)\geq0 from here .
Thanks for any help .
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  #2  
Old January 5th, 2009, 07:27 AM
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Originally Posted by mathaddict View Post
Given that (x-3) and (2x+1) are factors of
f(x)=ax^4+bx^3+13x^2=30x+9, find the values of a and b . With these values of a and b , show that f(x)\geq0 for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

4x^4-20x^3+13x^2+30x+9
(x-3)(2x+1)(2x^2-5x-3)

I am wondering how can i show that f(x)\geq0 from here .
Thanks for any help .
I suggest you find the coordinates and nature of the turning points of y = f(x) (note that 16x^3 -60x^2 + 26x + 30 = 2(x - 3)(2x + 1)(4x-5) .... )
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Old January 5th, 2009, 10:15 AM
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I think f(x) not >=0 with all real numbers .Maybe the problem is wrong .
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Old January 5th, 2009, 01:30 PM
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I think f(x) not >=0 with all real numbers .Maybe the problem is wrong .
You're wrong. On what grounds do you make this statement?
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Old January 5th, 2009, 02:21 PM
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Originally Posted by mathaddict View Post
Given that (x-3) and (2x+1) are factors of
f(x)=ax^4+bx^3+13x^2=30x+9, find the values of a and b . With these values of a and b , show that f(x)\geq0 for all x belongs to real numbers .

My working :

i found that a = 4 and b = -20

4x^4-20x^3+13x^2+30x+9
(x-3)(2x+1)(2x^2-5x-3)

I am wondering how can i show that f(x)\geq0 from here .
Thanks for any help .
Try factorising 2x^2-5x-3.
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Old January 5th, 2009, 04:16 PM
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Try factorising 2x^2-5x-3.
Well, you only do that if you want to do it the easy way .....
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Old January 5th, 2009, 04:30 PM
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Quote:
Originally Posted by mr fantastic View Post
You're wrong. On what grounds do you make this statement?
After factorising . I have f(x)=(x-3).(2x+1).(2x-3).(x+1)
If x belongs (3/2,3) ,f(x) <0 .Example x=2 ,I have f(2)=(2-3).(2.2+1).(2.2-3).(2+1)=(-1).5.1.3=-15
I thinks so that .
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Old January 5th, 2009, 05:04 PM
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Originally Posted by fanfan1609 View Post
After factorising . I have f(x)=(x-3).(2x+1).(2x-3).(x+1)
If x belongs (3/2,3) ,f(x) <0 .Example x=2 ,I have f(2)=(2-3).(2.2+1).(2.2-3).(2+1)=(-1).5.1.3=-15
I thinks so that .
2x^2 - 5x - 3 \neq (2x - 3)(x+1).

If you had expanded as a check you would have noticed this.
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  #9  
Old January 5th, 2009, 08:43 PM
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Quote:
Originally Posted by mr fantastic View Post
2x^2 - 5x - 3 \neq (2x - 3)(x+1).

If you had expanded as a check you would have noticed this.
Oh.I had a stupid mistake .Thanks for your check
2x^2 -5x -3 = (2x+1)(x-3) If that ,f(x)=(2x+1)^2(x-3)^2 .so that f(x) >= 0 with all x belongs real numbers .
Sorry for my previous comments.Thanks
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