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September 30, 2014

September 30, 2014

Posted by **anon** on Tuesday, December 20, 2011 at 10:29pm.

x^4 + 4x^3 - 9x^2 - 36x < 0

- algebra -
**Reiny**, Wednesday, December 21, 2011 at 9:06amx(x^3 + 4x^2 - 9x - 36) < 0

x( x^2(x+4) - 9(x+4) ) = 0

x(x+4)(x^2-9) < 0

x(x+4)(x+3)(x-3) < 0

so your critical values are

x = 0, -3, -4, and 3

Knowing that y = x^4 + .... is a quartic that rises in the 1st and 2nd quadrant because of the +1x^4 term, and it has 4 x-intercepts, you can make a rough sketch showing a W shape with those intercepts.

It should be obvious that the graph is below the x-axis ( y < 0 ) for values between -3 and -4 or between 0 and 3

so in standard notation:

-4 < x < -3 OR 0 < x < 3

I will leave it up to you to express it in interval notation

- algebra -
**anon**, Wednesday, December 21, 2011 at 10:36amThank you!

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