Solve Inequality: x^4+6x^3+7x^2-6x-8

[Sumedh]

Homework Statement

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

The Attempt at a Solution

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??

 

[ehild]

Do not open the brackets .
Investigate the signs of both factors instead.

ehild

 

[Tomer]

Do not open the brackets .
Investigate the signs of both factors instead.

ehild

Ehlid, did you notice there's a 5 on the right side of the equation?
I can't see how analyzing the signs of the expressions could help.

I'll think of that, I can't see an obvious answer.

 

[vela]

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

Yes, that's correct. Now factor the polynomial.

 

[Tomer]

I did find a solution, but I find it rather hard.

Do you know this theorem?
"
The integer root theorem. If an integer is a root of a polynomial whose coefficients are integers and whose leading coefficient is ±1, then that integer is a factor of the constant term."

Meaning: in the inequality you found, if there's a root that's an integer, it'll be either 1,-1,2,-2,4,-4,8 or -8.
A quick attempt or intuition will show you the correct one.

Do you know how to take it onward? You'll have to later use the same trick again...

Let us know if you're stuck!

If anyone found a better solution, please... :-)

 

[wukunlin]

very interesting little problem, excellent for any high schhol teachers who loves to induce unnecessary headaches

I did find a solution, but I find it rather hard.

Do you know this theorem?
"
The integer root theorem. If an integer is a root of a polynomial whose coefficients are integers and whose leading coefficient is ±1, then that integer is a factor of the constant term."

Meaning: in the inequality you found, if there's a root that's an integer, it'll be either 1,-1,2,-2,4,-4,8 or -8.
A quick attempt or intuition will show you the correct one.

Do you know how to take it onward? You'll have to later use the same trick again...

Let us know if you're stuck!

If anyone found a better solution, please... :-)

i solved it in 6 lines, all steps are doable to people who knows how to expand (x+a)(x+b) or factorize a binomial term into that form.

hint: treat (x^2+3x) as a single variable. eg say y = x^2+3x and reread the problem after you substitute the y in

 

[Tomer]

very interesting little problem, excellent for any high schhol teachers who loves to induce unnecessary headaches



i solved it in 6 lines, all steps are doable to people who knows how to expand (x+a)(x+b) or factorize a binomial term into that form.

hint: treat (x^2+3x) as a single variable. eg say y = x^2+3x and reread the problem after you substitute the y in

That's indeed a good idea, and I figured there's an easier way that I've missed.

 

[ehild]

Ehlid, did you notice there's a 5 on the right side of the equation?

Uhh. I did nor notice. But considering x²+3x a single variable as Wukunlin suggested is very promising...Try it.

ehild

 

[SammyS]

Homework Statement

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

The Attempt at a Solution

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??

The polynomial, $x^4+6x^3+7x^2-6x-8$, can be factored. x = 1 is a root, so (x-1) is a factor. Use synthetic division or long division to find the other factor (a 3rd degree polynomial). Then continue to factor the 3rd degree polynomial.

... But there's an easier way (IMO):

Notice that you can write the left hand side of the original inequality in the form of a sum times a difference, (y+a)(y-a), because:
x²+3x+1 = (x²+3x-1)+2
and x²+3x-3 = (x²+3x-1)-2 .

The original ]inequality becomes $[(x^2+3x-1)+2]\cdot[(x^2+3x-1)-2]\ge 5$

$(x^2+3x-1)^2-4\ge5\,.$ (Don't multiply out to get rid of the parentheses.)

Subtract 5 (a constant) from both sides and notice a new difference of squares on the left hand side of the resulting inequality. Factor the difference of squares.

It should be relatively easy from this point on, because you now have zero on the right, and the product of two quadratics on the left.​

Now, let's see if Sumedh shows up to respond to any of these posts ...

 

[Sumedh]

I am sorry for not responding quickly.

I was trying to do with the last method(#9) before it was posted,
but i failed to do, so i solved it with long division method and I got it.
Thank you very much.