# Manhattan GMAT Challenge Problem of the Week – 14 Feb 2011

*by*Manhattan Prep, Feb 14, 2011

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## Question

Which of the following represents the complete range ofxover which[pmath]x^3[/pmath]4[pmath]x^5[/pmath]< 0?(A) 0 < |x| <

(B) |x| >

(C) < x < 0 or < x

(D) x < or 0 < x <

(E) x < or x > 0

## Answer

One way to attack this problem is to factor the given expression:

*[pmath]x^3[/pmath]* 4*[pmath]x^5[/pmath]* = *[pmath]x^3[/pmath]*(1 4*[pmath]x^2[/pmath]*)

Notice that 1 4*[pmath]x^2[/pmath]* is a difference of squares. This part of the expression factors into

(1 2*x*)(1 + 2*x*)

So the whole expression becomes

*[pmath]x^3[/pmath]*(1 2*x*)(1 + 2*x*).

To trace the sign changes of the whole expression, track what happens to each part of the product.

*[pmath]x^3[/pmath]*is negative when*x*is less than zero, but its positive when*x*is greater than zero.- (1 2
*x*) is positive when*x*is less than , but its negative when*x*is greater than . (Careful about the sign change.) - (1 + 2
*x*) is negative when x is less than , but its positive when*x*> .

Now you have three break points where signs change: , 0, . This means that you have four regions to examine. You might set up a quick table to take care of the cases, or you can just talk your way through them.

1) *x* is less than : first term is negative, second is positive, third is negative, so the product is positive.

2) *x* is between and 0: first term is still negative, second is still positive, but third is now positive. So the product is negative.

3) *x* is between 0 and : first term is now positive. Second is still positive, third is positive, so the product is positive.

4) *x* is greater than : first term is positive, but now second term is negative. Third is still positive, so the product is negative.

Cases 2 and 4 give us a negative product. You can also test numbers, of course, but given the high powers, you might not want to raise fractions to these powers.

**The correct answer is C.**

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