Factorize a third degree polynomial
I'm currently trying to solve a problem which asks if a 3x3 matrix is diagonalizable, I know the method but when it comes to finding the roots, I have a third degree polynomial and I don't know how to factorize it to get the eigenvalues associated.
All the solutions on the internet and here about factorizing third degree polynomial are about specific case/obvious solutions and does not give a clear method like the method to factorize a second degree polynomial with steps.
Could you please provide me a method to find roots in every third degree polynomial?
If not this is the polynomial I found that I need to factorize : $X^3 - 3X - 2$
Thank you for taking your time to read my problem.
$\endgroup$ 15 Answers
$\begingroup$Try to "guess" some rational root $\;\cfrac rs\;$ , which by the Rational Root Theorem must fulfill $\;r\,\mid\,-2\;,\;\;s\,\mid\,1\;$ , and indeed $\;2\;$ is a root, so divide by $\;x-2\;$ :
$$x^3-3x-2=(x-2)(x^2+2x+1)=(x-2)(x+1)^2$$
and you have one simple root and one double one.
If there is no rational root then the task is much, but really much harder in the general case
$\endgroup$ $\begingroup$I can tell you how to factorise a Cubic polynomial. This would be a long lecture, so after reading this you try out with some polynomials.
Let's Start:
A third degree Polynomial is in the form of $$x^3 + bx^2+cx+d$$
Let the roots be $\alpha,\beta,\gamma$
Do you Know the symmetric notation:
$$ x^3 + (\sum_{}^{} \alpha )x^2 + (\sum_{}^{} \alpha\beta )x + \alpha\beta\gamma $$
Here,
$$\sum_{}^{} \alpha = \alpha+\beta+\gamma$$ $$\sum_{}^{} \alpha\beta = \alpha\beta + \beta\gamma +\gamma \alpha$$ $$\alpha\beta\gamma$$
What I am doing here is I am just expressing the coefficients in terms of roots just as we do in Quadratic Equation: $x^2+ (\alpha\beta)x+\alpha\beta$.
Or You can write ie like this also:
$$ x^3 + (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta\gamma +\gamma \alpha )x + \alpha\beta\gamma $$.
What can we get from these expressions??
We see that our $d$ is product of all roots, $b$ is Sum of all roots.
So, you can take these coefficients and try to guess the values.
I shall illustate this. In your case:
$$x^3 - 3x - 2$$
$$x^3 +0x^2- 3x - 2$$.
Here,
$$\alpha\beta\gamma = -2$$
Factors of -2 are $\pm1,\pm2$
Try to Guess here:
$$(-2)\cdot(+1)\cdot(+1)=-1$$ $$(-2)+(+1)+(+1)=0=\alpha+\beta+\gamma$$
Try for:
$$\alpha\beta + \beta\gamma +\gamma \alpha= (-2)(+1)+(+1)(+1)+(+1)(-2)=-2 + 1 +(-2) = -3 $$
Then You got all your zeroes:
Now Put them back :
$$(x-2)(x+1)(x+1)$$
Hence You got it.
Note: We are not changing any signs here because we have not changed any signs in the symmetric notation.
We give some work to "intutition" here!!
If there is existing $a$ here, then you divide that :
$$x^3+\frac{b}{a} x^2+\frac{c}{a} x + \frac{d}{a}$$
Or I say you use THE CUBIC FORMULA.
These seem scary. There are some books which say how to solve using cubic formula.
I use the above method.
Hope this helps!!
$\endgroup$ $\begingroup$There is no easy general method to factorize a third degree polynomial.
However in your case, you can notice that $2$ is a root of your polynomial :
$$2^3-3\times 2-2=0.$$
So you get
$$X^3-3X-2=(X-2)(aX^2+bX+c),$$
you develop and by identification you get
$$X^3-3X-2=(X-2)(X+1)^2.$$
$\endgroup$ $\begingroup$I will try our the factor of $2$.
Since $2^3-3(2)-2=0$, $x-2$ is a factor.
Since $(-1)^3-3(-1)-2=0$, $x+1$ is another factor.
Hence I will think of factorizing it as
$$x^3-3x-2=(x+1)(x-2)(x+c)$$
$$-2=1(-2)c$$
Hence $c=1$
$$x^3-3x-2=(x+1)^2(x-2)$$
$\endgroup$ $\begingroup$The general case of factoring a polynomial of degree 3 is quite painful. But in cases encountered in homework/assignements, you can usually:
- find an obvious root (try 0, 1, -1, i, -i)
- recognize some patterns (see for instance this example)
- use wolfram alpha
