Solving the Unsolvable: x^4+4x^3+14x^2=-4x-13

[thomasrules]

What are the ROots:

x^4+4x^3+14x^2=-4x-13

ok the 13 really causes a problem because you can't factor that.
So I move the right side to the left and then you can't find a number that fits so that it equals zero so I tried factoring it somehow but can't do it can someone help?

 

[Swapnil]

Well, if you look at the equation then you see that all the terms are positive. Thus, the graph of the function is a parabola which never touches the x-axis. Hence, there are no REAL roots and thus there are four complex roots.

I have no idea how to find those 4 roots, though..

 

[CRGreathouse]

Try to find a solution to $(x^2+ax+13)(x^2+bx+1)$. I chose this form because it automatically fits the first and last terms of your polynomial $x^4+4x^3+14x^2+4x+13$.

 

[thomasrules]

where did u pick up that equation from...

 

[Data]

Well, if you look at the equation then you see that all the terms are positive. Thus, the graph of the function is a parabola which never touches the x-axis. Hence, there are no REAL roots and thus there are four complex roots.

I have no idea how to find those 4 roots, though..

Well, $x^3$ and $x$ are hardly positive when $x<0$. But you are right, all the roots are complex here (the constant 13 is enough to keep it positive. However if the constant term were a small enough positive number [say, 0.1 instead of 13] you'd have real roots).

 

[thomasrules]

help me though :D

 

[Data]

I've already helped you in the identical thread in the homework help section :tongue2:

 

[thomasrules]

THanks I got the answer my friend mr data. I'm so smart :)

YOu caught me :D

 

[thomasrules]

here is another pickle...

Find Real ROots

$$y=x^4-4x^2-9x+36$$

AND NO I"M NOT ASKING YOU TO DO MY HOMEWORK, I TRIED THESE DAMN THINGS grrrrr

 

[CRGreathouse]

$$y=x^4-4x^2-9x+36$$

If you split the polynomial into $(ax^2+bx+c)(dx^2+ex+f)$ and simlify, you'l have a simple system that will tell you what the coefficints need to be. In fact, you can see that a and d in this example are 1, since x^4 appears with coefficient 1. Then you have

$$x^4-4x^2-9x+36=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(a+c+ac)x^2+(ad+bc)x+bd$$

Set each coefficient equal to the known coefficients:

$$a+c=0$$
$$a+c+ac=-4$$
$$ad+bc=-9$$
$$bd=36$$

and solve.

 

[shmoe]

$$x^4-4x^2-9x+36=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(a+c+ac)x^2+(ad+bc)x+bd$$

The x^2 coefficient is off.

 

[Swapnil]

$$y=x^4-4x^2-9x+36$$

If you split the polynomial into $(ax^2+bx+c)(dx^2+ex+f)$ and simlify, you'l have a simple system that will tell you what the coefficints need to be. In fact, you can see that a and d in this example are 1, since x^4 appears with coefficient 1. Then you have

$$x^4-4x^2-9x+36=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(a+c+ac)x^2+(ad+bc)x+bd$$

Set each coefficient equal to the known coefficients:

$$a+c=0$$
$$a+c+ac=-4$$
$$ad+bc=-9$$
$$bd=36$$

and solve.

Would this trick always work for any 4th degree polynomial? 'Cause if it does then it would be extremely helpful, because I always cry like a baby whenever I am asked to find the roots of a 4th degree by hand...

 

[CRGreathouse]

Would this trick always work for any 4th degree polynomial? 'Cause if it does then it would be extremely helpful ,because I cry whenever I am asked to find the roots of a 4th degree by hand...

Yes, always. You can't assume the x^2 coefficients are 1 unless the coefficient of x^4 is 1, though. (You can diide through to make this true, although youll have to multiply it back in the end.)

 

[CRGreathouse]

The x^2 coefficient is off.

Sorry. I wish I could claim that was "just so Thomas would do the work himself", but I actually just goofed. In any case, I'm sure thomasrules can work out the correct coefficient himself.

 

[thomasrules]

yes thanks guys, lol u think I'm taking advantage?

 

[thomasrules]

K though I wasn't taught that formula so...yea...

must be another way

 

[Swapnil]

Since we are on the topic of factoring and finding roots, I have a question. What ways are there to factor a polynomial which is greater than 2nd degree? I know that you can use grouping or use the rational root theorem to find all the possible rational roots and then use synthetic division to see if any of those are the root. Are there any other ways (which don't involve the use of computers or sucessive approximations)?

Also, isn't there a theorem that makes it impossible to factor a polynomial of degree > 5 or something like that?? (I think I have heard it someplace).