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December 21, 2014

December 21, 2014

Posted by **Alissa** on Tuesday, December 20, 2011 at 8:02pm.

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

- Algebra -
**Writeacher**, Tuesday, December 20, 2011 at 8:04pmAssistance needed.

- Alegra -
**Alissa**, Tuesday, December 20, 2011 at 9:32pmAh yeah which is why I post the question.

- Alegra -
**Ms. Sue**, Tuesday, December 20, 2011 at 9:47pmAlissa -- Writeacher corrected your spelling of "Algebra" so that a math expert would see your post.

Jiskha doesn't have any experts in Alegra (whatever that is).

- Alegra -
**Steve**, Wednesday, December 21, 2011 at 11:05amx^4 + 4x^3 - 9x^2 - 36x

= x(x^3 + 4x^2 - 9x - 36)

Cubics are hard to solve, so look for easy roots. They will be factors of 36.

A little synthetic division reveals

(x+4)(x+3)x(x-3)

Now since everything is single roots, the graph alternates between positive and negative between roots.

Since it's a quartic with a positive coefficient, it will be positive before the lowest root and after the highest root.

So,

y > 0 for x < -4

y < 0 for -4 < x < -3

y > 0 for -3 < x < 0

y < 0 for 0 < x < 3

y > 0 for x > 3

- Alegra -
**Alissa**, Wednesday, December 21, 2011 at 3:02pmThank you I appreciate your help and the fact that you didn't make a big issue out of a typo. Again thank you.

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