How Do You Factor a Quartic Polynomial into Quadratics?

[agoogler]

Homework Statement

p(x) = x^4+10x^3+26x^2+10x+1
p(x) = a(x)b(x) where a(x) and b(x) are quadratic polynomials with integer coefficients. It is given that b(1) > a(1). Find a(3) + b(2).

Homework Equations

p(x) = x^4+10x^3+26x^2+10x+1

The Attempt at a Solution

I tried to factor the given quartic but failed . Then I used wolfram-alpha and found that the factors are -
a](1+4x+x^2) and b](1+6x+x^2)
Now the problem is really easy to solve after this ( just evaluating a] at 3 and b] at 2 and adding the results equals 39 which is the correct answer ) , But I want to know how to factor that quartic into two quadratics . Please give hints.

 

[mfb]

If you look at the product of (c+bx+ax^2) and (e+dx+x^2) and compare it to your original polynomial, you can fix some factors quickly. The others are the result of calculations.

 

[haruspex]

If you look at the product of (c+bx+ax^2) and (e+dx+x^2) and compare it to your original polynomial, you can fix some factors quickly. The others are the result of calculations.

That's true, mfb, but when you see e.g. that c*e = 1 you have to conclude that c = e = ±1. c = 2, e = 1/2 isn't going to work, in my experience. But I don't know of a sound basis for that. Do you?

 

[agoogler]

If you look at the product of (c+bx+ax^2) and (e+dx+x^2) and compare it to your original polynomial, you can fix some factors quickly. The others are the result of calculations.

Awesome idea! Thanks.
Simplifying (c+bx+ax^2)(e+dx+fx^2)
I get (ce) + (cd+be)x + (bd+ae+cf)x^2 + (ad+bf)x^3 + (af)x^4

Now comparing the coefficients with x^4+10x^3+26x^2+10x+1
(ce) =1 (the constant term) Now if you assume the coefficients will be positive integers then both c and e =1 .
Also cd+be=10 which means d+b=10 ( since c and e =1 )
From the coefficient of x^4 , both a and f =1 .
Then from coefficient of x^2 we get bd+ae+cf=26 , but since a,e,c,f=1 we get bd=26-2=24.
Now we just have to find two numbers b and d whose product is 24 and sum is 10 .
So the answer is b=6 , d=4 and factors are -
a](1+4x+x^2) and b](1+6x+x^2)

Again thanks a lot , but anyone knows how to figure out if the coefficients (of factors) will be integers or not ?

 

[agoogler]

That's true, mfb, but when you see e.g. that c*e = 1 you have to conclude that c = e = ±1. c = 2, e = 1/2 isn't going to work, in my experience. But I don't know of a sound basis for that. Do you?

Yeah , that's what I also want to know.

 

[verty]

I think this problem is as difficult as finding roots of a quartic polynomial. In the case of a cubic polynomial, we want to factor $x^3 + bx^2 + cx + d$ into $(x^2 + px + q)(x + r)$, but this is just the same as finding the root x = -r and doing the division, and we know finding the root is difficult.

In the quartic case, if there was an easy method to factor the quartic into quadratics, we could find the roots by solving the quadratics. But it is not easy to find the roots, so it must be difficult to factor the quartic polynomial.

 

[haruspex]

I think this problem is as difficult as finding roots of a quartic polynomial. In the case of a cubic polynomial, we want to factor $x^3 + bx^2 + cx + d$ into $(x^2 + px + q)(x + r)$, but this is just the same as finding the root x = -r and doing the division, and we know finding the root is difficult.

In the quartic case, if there was an easy method to factor the quartic into quadratics, we could find the roots by solving the quadratics. But it is not easy to find the roots, so it must be difficult to factor the quartic polynomial.

That's not quite what I meant. If tasked with factoring a quartic with integer coefficients into two quadratics, it may be reasonable to assume that the problem poser has been kind enough that the factor polynomials will have rational coefficients. But does it then follow that they will have integer coefficients?

 

[mfb]

If there is a solution with rational numbers, then there is a solution with integers.
This is a result of Gauß' lemma.

Apart from that, integers are always the first thing one should try.

From the coefficient of x^4 , both a and f =1 .

You do not need f (and I did not include one in my post) - you can scale (c+bx+ax^2) by an arbitrary factor, so you don't need the same option for the second factor.

 

[agoogler]

If there is a solution with rational numbers, then there is a solution with integers.
This is a result of Gauß' lemma.

Apart from that, integers are always the first thing one should try.


You do not need f (and I did not include one in my post) - you can scale (c+bx+ax^2) by an arbitrary factor, so you don't need the same option for the second factor.

Oh , thanks. I didn't understand that before and thought that you missed it.

 

[Mentallic]

That's true, mfb, but when you see e.g. that c*e = 1 you have to conclude that c = e = ±1. c = 2, e = 1/2 isn't going to work, in my experience. But I don't know of a sound basis for that. Do you?

If the solution were to be

(x^2+ax+c)(x^2+bx+1/c)

where c\neq 0,1 then expanding this would yield

x^4+(a+b)x^3+(c+1/c+ab)x^2+(bc+a/c)x+1

Now, we're particularly interested in the coefficients of x^2 and x. Since we assumed that c\neq 1, then either

c+1/c+ab

is not an integer, or

bc+a/c

is not an integer (or both aren't), and since our quartic has all integer coefficients, this means our assumption of the value c can take is wrong, hence it must be 1.

 

[epenguin]

Before I say more there anything special you notice about this polynomial?

 

[agoogler]

Before I say more there anything special you notice about this polynomial?

Um , maybe the coefficients of both x and x^3 are same?
I can't think of anything else.
What is special about it?

 

[epenguin]

The completely general unspecial quartic has 5 coefficients, you could say, so it's not just a₁ and a₃ being equal that make this one special, also a₀ = a₄.

Divide by x², group, and see if that suggests something you could express in solvable form.

However it should be useful to you to check whether there was anything relevant in your book or course handouts etc. that you could have used.

 

[agoogler]

The completely general unspecial quartic has 5 coefficients, you could say, so it's not just a₁ and a₃ being equal that make this one special, also a₀ = a₄.

Divide by x², group, and see if that suggests something you could express in solvable form.

However it should be useful to you to check whether there was anything relevant in your book or course handouts etc. that you could have used.

Actually this is not a Homework question , I just found it somewhere and tried to solve it.

Okay , dividing by x^2 I get (x^2)+(10x)+(26)+(10/x)+(1/x^2).
Now , I've no idea how should I group this.

 

[epenguin]

Actually this is not a Homework question , I just found it somewhere and tried to solve it.

Okay , dividing by x^2 I get (x^2)+(10x)+(26)+(10/x)+(1/x^2).
Now , I've no idea how should I group this.

OK, it is more suggestive if we had had algebraic coeffients, in whivch case we would have had the form

x² + ax + b + a(1/x) + 1/x²

and you group by like coefficients.

 

[agoogler]

OK, it is more suggestive if we had had algebraic coeffients, in whivch case we would have had the form

x² + ax + bx² + a(1/x) + 1/x²

and you group by like coefficients.

Can you please explain the term bx^2? I guess it should be just b.
Can I group like this - a(x+1/x)+ x^2 + b + 1/x^2 ?
How will that benefit me?

 

[epenguin]

Can you please explain the term bx^2? I guess it should be just b.
Can I group like this - a(x+1/x)+ x^2 + b + 1/x^2 ?
How will that benefit me?

Right - just b, original now corrected.

So I should have said the algebraic form more completely, though I pointed out above that a₀ = a₄. This is preventing you recognising that 1 is a coefficient of two terms.

So you have got

(x² + 1/x²) + 10(x + 1/x) + 26

You can express the first bracket in terms of the second.

 

[Office_Shredder]

Bonus points if anyone can solve the problem without calculating what a(x) and b(x) are (I make no guarantee that it's possible but I feel like there's always some crazy trick with polynomials)

 

[haruspex]

(x² + 1/x²) + 10(x + 1/x) + 26

You can express the first bracket in terms of the second.

More generally, see Property 8 at http://en.wikipedia.org/wiki/Palindromic_polynomial

 

[agoogler]

Right - just b, original now corrected.

So I should have said the algebraic form more completely, though I pointed out above that a₀ = a₄. This is preventing you recognising that 1 is a coefficient of two terms.

So you have got

(x² + 1/x²) + 10(x + 1/x) + 26

You can express the first bracket in terms of the second.

Let me try.
(x² + 1/x²) + 10(x + 1/x) + 26
= ((x+1/x)²-2) + 10(x + 1/x) + 26
=(x+1/x)²+ 10(x + 1/x)+24

Am I right? Also what to do now? ( Sorry for asking you so many questions , but I'm very curious)

 

[agoogler]

Bonus points if anyone can solve the problem without calculating what a(x) and b(x) are (I make no guarantee that it's possible but I feel like there's always some crazy trick with polynomials)

Haha , yes there maybe some trick.

 

[epenguin]

Let me try.
(x² + 1/x²) + 10(x + 1/x) + 26
= ((x+1/x)²-2) + 10(x + 1/x) + 26
=(x+1/x)²+ 10(x + 1/x)+24

Am I right? Also what to do now? ( Sorry for asking you so many questions , but I'm very curious)

You have got a quadratic equation in a new variable, (x + 1/x) - call it z or whatever you like.

 

[agoogler]

You have got a quadratic equation in a new variable, (x + 1/x) - call it z or whatever you like.

Okay x+1/x =y
y^2+10y+24
= (y+6)(y+4)
which means y=-6,-4
x+1/x=-6
x^2+1=-6x
x^2+6x+1=0
Or x+1/x=-4
x^2+4x+1=0
So these are the two factors.
But what happens to the x^2 that we divided? Also please explain the "why" of the process.

 

[epenguin]

Okay x+1/x =y
y^2+10y+24
= (y+6)(y+4)
which means y=-6,-4
x+1/x=-6
x^2+1=-6x
x^2+6x+1=0
Or x+1/x=-4
x^2+4x+1=0
So these are the two factors.
But what happens to the x^2 that we divided? Also please explain the "why" of the process.

And if desired you can find the roots of the original equation by solving those two quadratics.
Nothing has to happen to the x², You can divide an equation by anything except 0 and the roots are the same.
I don't know about explaining the why, it kind of explains itself. See also haruspex' link.
To note
that you might not recognise a reiprocal polynomial immediately since you can multiply it by a constant, and also change the variable x into say x' = kx making something looking different, but you can work out what you get;
that every 4th degree polynomial can be made, by a linear substitution, into a reciprocal polynomial and hence solved in the way you did. One of the several ways of solving the quartic. The bad news is that in general to get the substitution involves solving a cubic equation, as does every other way of solving.

Best book I know on this area, old, cheap, probably online is Burnside and Panton, Theory of Equations,

 

[agoogler]

And if desired you can find the roots of the original equation by solving those two quadratics.
Nothing has to happen to the x², You can divide an equation by anything except 0 and the roots are the same.
I don't know about explaining the why, it kind of explains itself. See also haruspex' link.
To note
that you might not recognise a reiprocal polynomial immediately since you can multiply it by a constant, and also change the variable x into say x' = kx making something looking different, but you can work out what you get;
that every 4th degree polynomial can be made, by a linear substitution, into a reciprocal polynomial and hence solved in the way you did. One of the several ways of solving the quartic. The bad news is that in general to get the substitution involves solving a cubic equation, as does every other way of solving.

Best book I know on this area, old, cheap, probably online is Burnside and Panton, Theory of Equations,

I'll definitely try to read that book.