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# Is (x - 4)(x - 3)(x + 2)(x + 1) > 0

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Manager
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Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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Updated on: 27 Sep 2013, 08:06
5
00:00

Difficulty:

55% (hard)

Question Stats:

61% (01:46) correct 39% (01:41) wrong based on 299 sessions

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Is (x - 4)(x - 3)(x + 2)(x + 1) > 0

(1) 3 > x
(2) x > -1

Originally posted by HKHR on 27 Sep 2013, 08:01.
Last edited by Bunuel on 27 Sep 2013, 08:06, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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27 Sep 2013, 08:18
5
4
Is (x - 4)(x - 3)(x + 2)(x + 1) > 0

Is $$(x + 2)(x + 1)(x - 3)(x - 4) > 0$$? --> the "roots" are -2, -1, 3, and 4 (equate each multiple to zero to get the roots and list them in ascending order), this gives us 5 ranges:

$$x<-2$$;
$$-2<x<-1$$;
$$-1<x<3$$;
$$3<x<4$$;
$$x>4$$.

Now, test some extreme value: for example if $$x$$ is very large number then all multiples ((x + 2), (x + 1), (x - 3), and (x - 4)) will be positive which gives the positive result for the whole expression, so when $$x>4$$ the expression is positive.

Now the trick: as in the 5th range expression is positive then in 4th it'll be negative, in 3rd positive, in 2nd negative, and finally in 1st it'll be positive again: + - + - +. So, the ranges when the expression is positive are: $$x<-2$$ (1st range), $$-1<x<3$$ (3rd range) and $$x>4$$ (5th range).

So, the question asks whether $$x<-2$$, $$-1<x<3$$ or $$x>4$$

(1) 3 > x. Not sufficient.
(2) x > -1. Not sufficient.

(1)+(2) -1<x<3. Sufficient.

Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html
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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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27 Sep 2013, 10:23
omg! Simple and nice solution that I understood. But, what is wrong with this:

(1) Since X<3, pick X = 2. Get (-2)*(-2)*(3)*(3) >0 ( for that matter any number or fraction less than 3 would give this expression >0) - Suff

(2) X>-1, Let x = 0 similar approach above, so is suff.

Seems to0 simple for this level of difficulty, but why is this wrong, what was not considered?!
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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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27 Sep 2013, 11:01
1
igotthis wrote:
omg! Simple and nice solution that I understood. But, what is wrong with this:

(1) Since X<3, pick X = 2. Get (-2)*(-2)*(3)*(3) >0 ( for that matter any number or fraction less than 3 would give this expression >0) - Suff

(2) X>-1, Let x = 0 similar approach above, so is suff.

Seems to0 simple for this level of difficulty, but why is this wrong, what was not considered?!

You cannot check sufficiency based only on one arbitrary value. Consider x=-1.5 for (1) and x=3.5 for (2).
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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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09 Dec 2014, 07:56
HKHR wrote:
Is (x - 4)(x - 3)(x + 2)(x + 1) > 0

(1) 3 > x
(2) x > -1

What Bunuel did may be the right way to solve this problem, but the easier way is to take some values and check

S1: Here X can be anything number less than 3 to 0 on the positive side and 0 to infinity on the negative side.

Lets check with 2,1,0..all of them give positive numbers. The fractions in between 0 & 3 also give positive numbers. Now -1 & -2 give 0 and -3,-4 etc give -ve numbers. So insufficient

S2: we have checked with -1, -2 etc. But lets check with fractions between -1 & 0 for eg -1/2 [which is greater than -1] gives a positive number. Let us also check for numbers equal to or greater than 3.... 3,4 give negative numbers and above four it is positive. So insufficient.

Combining 1 & 2, for the checks done above, we can easily say that 3>X>-1 and hence Answer C.

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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0  [#permalink]

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03 Nov 2018, 08:10
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Re: Is (x - 4)(x - 3)(x + 2)(x + 1) > 0   [#permalink] 03 Nov 2018, 08:10
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