Solve for \(x\).

- mathmath333

Solve for \(x\).

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- mathmath333

\(\large \color{black}{\begin{align} (x^2+3x+1)(x^2+3x-3)\geq 5\hspace{.33em}\\~\\
\end{align}}\)

- Loser66

.

- ganeshie8

One dumb method is to expand everything out and factor it again absorbing that 5 on right hand side

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ybarrap

And that gives
$$
(x-1) (x+1) (x+2) (x+4)-5
$$
Then find for what values of x is this greater or equal to 0

- ganeshie8

Pretty sure you mean
$$
(x-1) (x+1) (x+2) (x+4)\ge 0
$$
Then find for what values of x is this greater or equal to 0

- IrishBoy123

.

- mathmath333

how did u factor that
\(\large \color{black}{\begin{align} (x-1) (x+1) (x+2) (x+4)\ge 0\hspace{.33em}\\~\\
\end{align}}\)

- dan815

x(x+3) +1
x(x+3) -3|dw:1436812802011:dw|

- dan815

|dw:1436812913492:dw|

- dan815

u know the rate of increase of both parabolas so u see the first instance they pass pr0duct 5

- dan815

then from axis of symmetry at 1.5 u can see the other end too

- anonymous

@mathmath333 $$(x^2+3x+1)(x^2+3x-3)-5=(u+4)u-5=u^2+4u-5=(u+5)(u-1)$$for \(u=x^2+3x-3\)... and then we see that \((x^2+3x-3+5)(x^2+3x^2-3-1)=(x^2+3x+2)(x^2+3x-4)\) and both of these factor easily $$x^2+3x+2=(x+1)(x+2)\\x^2+3x-4=(x+4)(x-1)$$

- anonymous

so for \(f(x)=(x+1)(x+2)(x+4)(x-1)\ge 0\) we know that the solution is going to be every other interval between the roots, so either $$(-\infty,-4]\cup[-2,-1]\cup[1,\infty)\\\text{or}\\ [-4,-2]\cup[-1,1]$$

- anonymous

to determine which just test a point in one of the intervals: $$f(0)=(1)(2)(4)(-1)=-8<0$$so it follows the second intervals solve \(f(x)\le 0\) while the first solve \(f(x)\ge 0 \)

- mathmath333

i d k why i am not able to see latex.

- dan815

reload

- anonymous

##### 1 Attachment

- mathmath333

reloaded twice.

- anonymous

well, you don't need to test both intervals; you already know that since each root has multiplicity 1 the function is crossing between each so alternating between positive and negative; then you just test a point in any one of the intervals to distinguish which of these two solutions is for nonnegative \(f\) and which is for nonpositive \(f\)

- Loser66

@oldrin.bataku Since I use table to determine it. It is like |dw:1436813611624:dw|

- Loser66

|dw:1436813832253:dw|

- anonymous

no I understand, I'm just saying that's more work than necessary, since you only need to check the sign of one of the intervals (not all of them) since we already determined the solution must either be \((-\infty,-4]\cup[-2,-1]\cup[1,\infty)\) or \([-4,-2]\cup[-1,1]\)

- anonymous

checking the sign on say \([-2,-1]\) is enough to tell you the sign on \((-\infty,-4]\cup[-2,-1]\cup[1,\infty)\) since the function is alternating in sign between intervals so it must have the same sign for all of these, and the opposite sign for \([-4,-2]\cup[-1,1]\)

- Loser66

Got you!!:)

- Loser66

@oldrin.bataku Is there any other way to solve it?

- ybarrap

The best way is the @oldrin.bataku approach, if you can see it.
Otherwise, (the hard way) is to us the Rational Root Theorem, which gives you the possible roots. After multiplying everything out we get:
$$
(x^2+3x+1)(x^2+3x-3)-5=x^4+6 x^3+7 x^2-6 x-8
$$
The possible roots are then, \(\pm1,\pm2,\pm4,\pm8\).
Then using synthetic division, test each.
https://en.wikipedia.org/wiki/Rational_root_theorem
https://en.wikipedia.org/wiki/Synthetic_division
I knew a most there was 1 positive root, using Descartes' rule of signs, that also helped -
https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
I got lucky trying x=1,-1 first then everything else was easier because then I was just left with an easy quadratic to factor.

- freckles

like the only other way I found is just a similar substitution but everything else is the same
choose u=x^2+3x makes things factorable over the integers
\[u=x^2+3x \\ (u+1)(u-3)-5 \ge 0 \\ (u^2-2u-3)-5 \ge 0 \\ u^2-2u-8 \ge 0 \\ (u-4)(u+2) \ge 0 \\ (x^2+3x-4)(x^2+3x+2) \ge 0 \\ (x-1)(x+4)(x+1)(x+2) \ge 0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.