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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 07:49
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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0?

(A) fewer than 5 (B) 5 (C) 6 (D) 7 (E) more than 7



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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 08:29
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Hi
The function can be re-written as; X^4-X^2-8X^3+8X or X^2*(X^2-1)-8X*(X^2-1) or (X^2-8X)*(X^2-1)<0

two cases
X^2-8X>0=> X*(X-8)>0 sol (8;+ inf) U (0;-inf)
X^2-1<0 sol (-1;1)
For the first case there is no integer solution

case 2
X^2-8X<0=>X*(X-8)<0 sol all integers exept 0 and 8
X^2-1>0 sol (1;+ inf) U (-1; - inf)

Think it is E)
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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 09:31
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The inequality reduces to
x(x-1)(x+1)(x-8) < 0

hence the solution set is
-1 < x < 0 and 1 < x < 8

so the total no of integers is 6 viz., {2, 3, 4, 5, 6, 7}

Hence (C)
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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 10:08
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Answer is 6, which is (C)
x^4 - 8*x^3 - x^2 + 8x
=x^3(x-8) -x(x-8)
=(x-8)*(x^3-x)

If x = 0 or 1 the result of the equation is 0
Any -ve value of x will result +ve value of the above equation.
For ex. x=-1, so (-1-8)*(-1-1)=18
or x=-2, so (-2-8)*(-8-2)=100

but if x is any value between 2 and 7 the result is <0
For ex. x=2, so (2-8)*(2^3-2)
For ex. x=7, so (7-8)*(7^3-7)

Factor one is -ve but factor two is positive, so the result is -ve
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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 16:13
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Yes... agree with (C) :)

f(x) = x^4-8x^3-x^2+8x
= x*(x^3 - 8*x^2 - x + 8)

1 is a root of x^3 - 8*x^2 - x + 8, thus we have:

x^3 - 8*x^2 - x + 8
= (x-1)*(x^2 +b*x -8)
= (x^3 +b*x^2 - 8*x -x^2 -b*x +8)

=> b-1 = -8
<=> b= -9

Then, we have:
f(x) = x*(x-1)*(x^2 -9*x -8)

-1 is root of x^2 -9*x -8, so we have:
f(x) = x*(x-1)*(x+1)(x-8)

Now, we want to know for which walues of x we have f(x) < 0. I suggest to build up a table of signs (Fig 1).

So, we can conclude that f(x) < 0 when -1 < x < 0 or 1 < x < 8. In terms of possible integers, we have thus 2, 3, 4, 5, 6 and 7.
Attachments

Fig1_Table of signs.gif
Fig1_Table of signs.gif [ 6.24 KiB | Viewed 1373 times ]

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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 17:13
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BG
Hi
The function can be re-written as; X^4-X^2-8X^3+8X or X^2*(X^2-1)-8X*(X^2-1) or (X^2-8X)*(X^2-1)<0>0=> X*(X-8)>0 sol (8;+ inf) U (0;-inf)
X^2-1<0 sol (-1;1)
For the first case there is no integer solution

case 2
X^2-8X<0>X*(X-8)<0>0 sol (1;+ inf) U (-1; - inf)

Think it is E)
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Hi BG,

My answer was E too and my reasoning was equal to yours.
But it seems that the answer is 6.
What is our mistake?
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If f(x)= x^4-8x^3-x^2+8x , for many integers x is f<0? 09 Mar 2007, 17:24
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BG
Hi
The function can be re-written as; X^4-X^2-8X^3+8X or X^2*(X^2-1)-8X*(X^2-1) or (X^2-8X)*(X^2-1)0=> X*(X-8)>0 sol (8;+ inf) U (0;-inf)
X^2-1X*(X-8)0 sol (1;+ inf) U (-1; - inf)

Think it is E)
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I got it I got it!
case 2
X^2-8XX*(X-8)<0 sol integers (0;8) <------ here is the mistake



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