Solve the inequality

**sumedh** · August 27th 2011, 12:37 AM

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??

**Siron** · August 27th 2011, 12:43 AM

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? ...

**FernandoRevilla** · August 27th 2011, 12:44 AM

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.

**Siron** · August 27th 2011, 12:51 AM

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.

**sumedh** · August 27th 2011, 01:04 AM

thank you i got the answer

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