# Factoring a quartic polynomial

#### Mr Davis 97

1. The problem statement, all variables and given/known data
Factor $x^4-3x^2+9$ over the reals

2. Relevant equations

3. The attempt at a solution
I am factoring this polynomial over the reals. So there are two options. It will either split into two linear factors and an irreducible quadratic, or two irreducible quadratics. I'm really not too sure how to proceed from here... Should I first start looking for a real root?

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#### andrewkirk

Homework Helper
Gold Member
The polynomial is easily factored over the complex numbers by first solving it as a quadratic in $x^2$. By then taking both square roots of both solutions, you'll have four complex roots of the quartic. The question then is whether any product of three or fewer degree-one polynomial factors each of the form $(x-r)$ where $r$ is a complex root, gives a polynomial with only real coefficients.

#### Mr Davis 97

The polynomial is easily factored over the complex numbers by first solving it as a quadratic in $x^2$. By then taking both square roots of both solutions, you'll have four complex roots of the quartic. The question then is whether any product of three or fewer degree-one polynomial factors each of the form $(x-r)$ where $r$ is a complex root, gives a polynomial with only real coefficients.
So I have $(x+\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x+\sqrt{\frac{3}{2}(1+i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1+i \sqrt{3})})$. This seems very messy. Is there any way to simplify it?

#### Mark44

Mentor
So I have $(x+\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x+\sqrt{\frac{3}{2}(1+i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1+i \sqrt{3})})$. This seems very messy. Is there any way to simplify it?
Looks fine to me. I don't think this factorization lends itself to much more simplification.

#### Mr Davis 97

Looks fine to me. I don't think this factorization lends itself to much more simplification.
So is there a way to get from this factorization to the product of two quadratics with real coefficients?

#### Ray Vickson

Homework Helper
Dearly Missed
So I have $(x+\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x+\sqrt{\frac{3}{2}(1+i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1+i \sqrt{3})})$. This seems very messy. Is there any way to simplify it?
The complex numbers $\sqrt{(3/2) (1\pm i \sqrt{3})}$ simplify a lot.

#### mathwonk

Homework Helper
it seems it shoud factor into two real quadratics, no?

#### andrewkirk

Homework Helper
Gold Member
So is there a way to get from this factorization to the product of two quadratics with real coefficients?
Yes there is. Start by writing the two roots of the quadratic in $x^2$ in polar form, which has been made easy by the specially selected coefficients given in the problem.

#### lurflurf

Homework Helper
we are looking for quadratic factors ie
$x^4-3x^2+3^2=(a x^2+b x+c)(d x^2+e x+f)$
we can reduce the choices by considering each coefficient in turn

x^4
a d =1
we know a and d since there are very few factorizations
x^3
a e+b d=0
we know b+e
x
b f+c e=0
we know c-f
x^0
c f=9
we know c and f
x^2
a f+b e+c d
we know b and e
we are done

note that two coefficients being zero and the remaining coefficients being small helps greatly

#### stevendaryl

Staff Emeritus
Note that if you multiply or add two numbers that are complex-conjugates, then the result is real:

$(a+ib)(a-ib) = a^2 + b^2$
$(a+ib) + (a-ib) = 2a$

So it follows that $(x + \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x + \sqrt{\frac{3}{2} (1-i\sqrt{3})})$ can be expressed using just real coefficients. Similarly for $(x -\sqrt{\frac{3}{2} (1+i\sqrt{3})})(x - \sqrt{\frac{3}{2} (1-i\sqrt{3})})$

So your real factorization is $[(x + \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x + \sqrt{\frac{3}{2} (1-i\sqrt{3})})] [(x - \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x - \sqrt{\frac{3}{2} (1-i\sqrt{3})})]$

There's a big simplification that comes from using the polar form of complex numbers: $a + ib = \sqrt{a^2 + b^2} e^{i \theta}$ where $\theta$ is chosen so that $cos(\theta) = \frac{a}{\sqrt{a^2+b^2}}$ and $sin(\theta) = \frac{b}{\sqrt{a^2+b^2}}$. Then the square root is easy:

$\sqrt{a+ib} = (a^2 + b^2)^{\frac{1}{4}} e^{i \frac{\theta}{2}}$

#### lurflurf

Homework Helper
So I have $(x+\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1-i \sqrt{3})})(x+\sqrt{\frac{3}{2}(1+i \sqrt{3})})(x-\sqrt{\frac{3}{2}(1+i \sqrt{3})})$. This seems very messy. Is there any way to simplify it?
notice
$x^4-3x^2+9=(x-a)(x-a^\ast)(x+a)(x+a^\ast) % =[ x^2-(a+a^\ast)x+a^\ast a ] [ x^2+(a+a^\ast)x+a^\ast a ]$
$=x^4-[(a+a^\ast)^2-2a^\ast a]x^2+(a^\ast a)^2$
It is then easy find a+a* and a*a to effect the factorization.
You could also find a in rectangular or polar coordinates if you wanted to.

#### Ray Vickson

Homework Helper
Dearly Missed
Note that if you multiply or add two numbers that are complex-conjugates, then the result is real:

$(a+ib)(a-ib) = a^2 + b^2$
$(a+ib) + (a-ib) = 2a$

So it follows that $(x + \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x + \sqrt{\frac{3}{2} (1-i\sqrt{3})})$ can be expressed using just real coefficients. Similarly for $(x -\sqrt{\frac{3}{2} (1+i\sqrt{3})})(x - \sqrt{\frac{3}{2} (1-i\sqrt{3})})$

So your real factorization is $[(x + \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x + \sqrt{\frac{3}{2} (1-i\sqrt{3})})] [(x - \sqrt{\frac{3}{2} (1+i\sqrt{3})})(x - \sqrt{\frac{3}{2} (1-i\sqrt{3})})]$

There's a big simplification that comes from using the polar form of complex numbers: $a + ib = \sqrt{a^2 + b^2} e^{i \theta}$ where $\theta$ is chosen so that $cos(\theta) = \frac{a}{\sqrt{a^2+b^2}}$ and $sin(\theta) = \frac{b}{\sqrt{a^2+b^2}}$. Then the square root is easy:

$\sqrt{a+ib} = (a^2 + b^2)^{\frac{1}{4}} e^{i \frac{\theta}{2}}$
That is what I hinted at in post #6, but have received no feedback from the OP to date. (Actually, maybe not so much of a hint; perhaps more of a pre-hint, but with the motivation to have the OP do some more work on the problem.)

#### ehild

Homework Helper
1. The problem statement, all variables and given/known data
Factor $x^4-3x^2+9$ over the reals

2. Relevant equations

3. The attempt at a solution
I am factoring this polynomial over the reals. So there are two options. It will either split into two linear factors and an irreducible quadratic, or two irreducible quadratics. I'm really not too sure how to proceed from here... Should I first start looking for a real root?
Note that the polynomial can be written as $(x^2+3)^2-9x^2$.

"Factoring a quartic polynomial"

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