I am suppose to graph f(x)=x^5-3x^3-x^2-4x-1 but I am having trouble finding its zeros by factoring it. Please help and thank you in advance ^_^

- anonymous

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- anonymous

\[x^5-3x^3=x^3(x-3)\\
x^5-3x^3-x^2-4x-1=x^3(x-3)-(x^2+4x+1)\]

- whpalmer4

\[x^5-3x^3-x^2-4x-1\]I'll give you a hint: 1 of the factors is one of (x+1), (x-1), (x^2+1), (x^2-1). I guessed this based on the fact that the constant term is -1 and the coefficient of x^5 is 1.
By Descartes' Rule of Signs, we'll have 1 positive root, and 0, 2, or 4 negative roots, and the rest will be complex conjugate pairs, 5 roots in all.

- whpalmer4

Nuts, I must have miscounted when doing DRoS...never mind that part!

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## More answers

- anonymous

I need to practice factoring. I can't get this one.

- whpalmer4

Did you try my suggestions?

- anonymous

Im trying. But cannot see it.

- whpalmer4

Okay, try dividing the polynomial by (x^2+1)

- anonymous

Polynomial division is the most hard thing to do on a text line.

- whpalmer4

That's what paper and pencil is for :-)

- anonymous

\[x^5 -3x^3 \div (x^2+1)=x^3\]
Rest is \(-2x^3\)
\[-2x^3-x^2-4x-1 \div (x^2+1)=-2x\]
And the Rest is \(-x^2-2x-1\)
\[-x^2-2x-1 \div(x^2+1)=-1\]
Rest is -2x
So we have:
\[x^5-3x^3-x^2-4x-1=\\
x^3(x^2+1)-2x(x^2+1)-(x^2+1) -2x=\\
f(x)=(x^3-2x-1)(x^2+1)-2x
\]
It didnt factor out correctly.

- anonymous

@whpalmer4 can you check it out if I divided it correctly?

- anonymous

Yeah I made a mistake somewhere. I will try to check it.

- whpalmer4

x^2+1 | x^5-3x^3-x^2-4x-1 first term will be x^3
-x^5-x^3
----------
-4x^3 - x^2 next term will be -4x
4x^3 +4x
----------------
-x^2 - 1 final term is -1
x^2 + 1
----------
0
(note, I prefer to multiply by -1 and add rather than subtract, if you're thinking this looks a bit weird)
so \((x^2+1)(x^3-4x-1)\) is my factoring.

- whpalmer4

\[(x^2+1)\] has solutions \(x = \pm i\). The other thing is a cubic, and the solutions are the usual wild cubic stuff...

- anonymous

The first division is wrong:
\[x^5 -3x^3 \div (x^2+1)=x^3\]
Should remainder should be \(-4x^3\) Thats where I made the mistake.

- anonymous

Its a very unusual problem if it takes solving cubic equations.

- whpalmer4

Oh, I see what you're doing with that notation...didn't catch on at first. I just divide the first terms

- whpalmer4

It's entirely possible we're supposed to be cleverer rather than brute-forcing it.

- whpalmer4

I always enjoy the OS experience when we spend considerable effort trying to solve what turns out to be something other than the intended problem :-)

- whpalmer4

The typing/notation mistakes never seem to give us problems that are easier than the original, either :-)

- anonymous

It can happen, specially when the user that submitted the question is not even trying.

- whpalmer4

especially frustrating is when you ask "okay, is this what it looks like" (after typesetting something that seems plausible), you work the problem for a while, and then having painted yourself into a corner, you realize that they didn't even look carefully enough before saying "yes"!

- anonymous

I don't think its a typo though. I think @madatmath didn't understand the problem right. He probably thinks he needs to find these solutions he actually doesnt.

- anonymous

... I am constantly reading your answers.. Sorry if I seem like I'm not trying but I don't really understand this problem overall so I'm just carefully watching your steps. I understand what you did so far and I'm grateful for you help and effort. ^_^

- whpalmer4

No, we're not disparaging you at all, @madatmath!

- anonymous

so I don't need to find the solution if I want to graph it?

- anonymous

Except I do think the problem was somehow misinterpreted. The solution up to now should be relevent.

- anonymous

It depends on what the problem explicitly stated, what topics are you studying and what course are you doing.

- whpalmer4

Also, what is the expected use of the graph? Just find the general shape and behavior, or make something so precise that someone can read off values rather than evaluating the polynomial? My guess is the former is more likely than the latter, but I'm not the one doing the homework or taking the class.

- anonymous

I throughly doubt some 7th grade algebra 2 problem would take solving cubic equations. Maybe some Math college algebra.

- whpalmer4

You can always do it by evaluating a bunch of points and plotting them.

- anonymous

this is summer homework for calculus class. the instructions states: "sketch the graphs of the following functions. For each graph you need to identify the following characteristics: (a) domain, (b) range, (c) x and y intercepts, (d) symmetry, (e) asymptotic behavior, & (f) a table of value"

- whpalmer4

The trick is having an idea about which points to do...I'll tell you this, it starts negative, crosses over the x-axis between x=-2 and x = -1, crosses back under the x-axis between x=-1 and x=0, then hook up over the x-axis (never again to be negative) just to the right of x=2...

- anonymous

Ok, now most of those items are pretty straightforward, domain, range, y intercepts, symetry and value table.

- anonymous

I think you can state the x intercepts just by saying that those are the solutions to the cubic equation we found:
\[x^3-4x-1=0\]

- anonymous

No need to actually solve it by pen and paper.

- whpalmer4

One thing to know with these higher order polynomials where the coefficients of the x^4, x^5, etc. terms aren't really small numbers (< 1) is that all of the "action" on the graph takes place between -10 < x < 10 or even smaller range, because otherwise, the value of x being multiplied over and over swamps the value of the coefficients and it's either really big or really small.

- whpalmer4

For example, here's a graph of your function, done without any thought to making what happens in the interesting region visible:

##### 1 Attachment

- anonymous

Sorry if I'm not getting this but I think what you're saying is that for this particular problem, we just need to factor it until there and there's no need to go any further, correct? As for plotting, I can just evaluating points.

- whpalmer4

See how quickly it "blows up"? You can barely tell that there might be something interesting between -3 < x < 3

- whpalmer4

Now here I've graphed the region I mentioned, just scaled it differently on the y-axis so you can actually see something!

##### 1 Attachment

- whpalmer4

Quite a difference, eh? :-)

- anonymous

a) b) Domain and range: \(\mathbb R\)
c) x intercepts are \(x \in \mathbb R :x^3-4x-1=0\)
y intercept is y=-1
d) There is no axis of symmetry.
e) There is no asymptotes.
f) f(-1)=?
f(0)=-1
f(1)=-9

- anonymous

Oh, aha, you didn't have to provide me with all the answers ^_^"... I just didn't know how to approach this problem, but thank you to both for clarifying things up and putting in so much effort into a problem that I made harder than it should be.

- anonymous

From the graph whpalmer showed you should also table f(-2), f(2) because they are very close to the actual solutions.

- anonymous

I'm really nervous for calculus class next year now T_T

- anonymous

Nah, don't be. You were on the right track. Sometimes we get hangup on some difficult thing that is not really there. But its always a chance to learn things. all that whpalmer4 explained are pretty interesting stuff to know for algebra and calculus.
That only happens if you take your time to study the problems.

- anonymous

agreed :) Thanks again!

- whpalmer4

Went off to make a snack and my computer lost its marbles while I was away :-(
I wouldn't sweat the calculus class too much for this reason — my calculus classes touched on stuff you could use for graphing, sure, but even before graphing calculators doing painstaking graphs of ugly functions wasn't a big emphasis. mostly we just did enough to get comfortable with using the "tools" calculus provided

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