1. Solve the inequality

Solve the inequation $\displaystyle (x^2+3x+1)(x^2+3x-3)\ge 5$

on opening the brackets i got

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

am i right??
after that what should i do??

2. Re: Solve the inequality

It looks fine, notice you can write $\displaystyle x^4+6x^3+7x^2-6x-8=(x-1)\left(ax^3+bx^2+cx+d)$
Determine $\displaystyle a,b,c,d$ by Horner's scheme or the euclidian division, afterwards do the same for $\displaystyle ax^3+bx^2+cx+d$ so you can write finally the function as:
$\displaystyle x^4+6x^3+7x^2-6x-8=(x-1)(x-a)(x-b)(x-c)$
Where you've determined $\displaystyle a,b,c$ and which is useful to make the sign-table.

Is this clear? ...

3. Re: Solve the inequality

Originally Posted by sumedh
am i right?? after that what should i do??
Right, now express $\displaystyle x^4+6x^3+7x^2-6x-8=(x-1)(x+1)(x+2)(x+4)$ .

Edited: Sorry, I didn't see Siron's post.

4. 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.

5. Re: Solve the inequality

thank you i got the answer