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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.

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    $\begingroup$ Note that $X+1$ divides your polynomial. $\endgroup$ – André Nicolas Jun 28 '16 at 16:30
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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

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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.

http://mathworld.wolfram.com/CubicFormula.html

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

These seem scary. There are some books which say how to solve using cubic formula.

I use the above method.

Hope this helps!!

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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

$$X^3-3X-2=(X-2)(X+1)^2.$$

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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 $c=1$

$$x^3-3x-2=(x+1)^2(x-2)$$

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The general case of factoring a polynomial of degree 3 is quite painful. But in cases encountered in homework/assignements, you can usually:

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