# Solve the inequality

• Aug 27th 2011, 12:37 AM
sumedh
Solve the inequality
Solve the inequation $(x^2+3x+1)(x^2+3x-3)\ge 5$

on opening the brackets i got

$x^2(x^2+3x-3)+3x(x^2+3x-3)+1(x^2+3x-3)\ge 5$
$x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5\ge 0$
$x^4+6x^3+7x^2-6x-8\ge 0$

am i right??
after that what should i do??
• Aug 27th 2011, 12:43 AM
Siron
Re: Solve the inequality
It looks fine, notice you can write $x^4+6x^3+7x^2-6x-8=(x-1)\left(ax^3+bx^2+cx+d)$
Determine $a,b,c,d$ by Horner's scheme or the euclidian division, afterwards do the same for $ax^3+bx^2+cx+d$ so you can write finally the function as:
$x^4+6x^3+7x^2-6x-8=(x-1)(x-a)(x-b)(x-c)$
Where you've determined $a,b,c$ and which is useful to make the sign-table.

Is this clear? ...
• Aug 27th 2011, 12:44 AM
FernandoRevilla
Re: Solve the inequality
Quote:

Originally Posted by sumedh
am i right?? after that what should i do??

Right, now express $x^4+6x^3+7x^2-6x-8=(x-1)(x+1)(x+2)(x+4)$ .

Edited: Sorry, I didn't see Siron's post.
• Aug 27th 2011, 12:51 AM
Siron
Re: Solve the inequality
I doens't matter :). But I thought the only difficulty of this exercice was the 4th degree inequality so sumedh could solve it by himself but offcourse he can decide what he thinks which is the best for himself.
• Aug 27th 2011, 01:04 AM
sumedh
Re: Solve the inequality
thank you i got the answer:)