How To Find Degree Of A Polynomial

Take an arbitrary polynomial of degree 3. If you mean finding it’s roots… first: you can divide it by the coefficient at [math]x^3[/math] to get an equation [math]x^3+ax^2+bx+c=0[/math]. Now substitute [math]t=x+\frac{a}{3}[/math] and the coefficient at x^2 will disappear and you will get [math]x^3+px+q=0[/math]. Suppose also that [math]p\neq 0,q\neq 0[/math] as otherwise the solutions are trivial.How to solve such an equation? I will show you so-called Cardano’s method:Suppose [math]x=\alpha+\beta[/math]. Substitute it to the equation and get:[math]\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)+p(\alpha+\beta)+q=0[/math][math]\alpha^3+\beta^3+(\alpha+\beta)(3\alpha\beta+p)+q=0[/math].Now here comes the trick: as we have a freedom choosing [math]\alpha[/math] and [math]\beta[/math] we shall set [math]3\alpha\beta+p=0[/math]. Now we’ve got a system:[math]\left\{\begin{array}{lll} \alpha^3+\beta^3+q&=&0\\\alpha\beta&=&-\frac{p}{3}\end{array}\right.\Leftrightarrow \left\{\begin{array}{lll} \alpha^6+q\alpha^3-\frac{p^3}{27}&=&0\\\beta&=&-\frac{p}{3\alpha}\end{array}\right.[/math]You’ve got a quadratic equation, solve it, find [math]\alpha,\beta[/math]. Sum them up and get x.

A monomial in [math]n[/math] variables [math]x_1,\ldots,x_n[/math] over a set [math]S[/math] [math]([/math]often [math]\mathbb Z[/math] or [math]\mathbb Q[/math] or [math]\mathbb R[/math] or [math]\mathbb C)[/math] is an expression of the form [math]c x_1^{e_1} \cdots x_n^{e_n}[/math], where each [math]e_i \in {\mathbb N} \cup \{0\}[/math] and [math]c \in S[/math].To each monomial is associated a nonnegative integer, called the degree of the monomial, and given by [math]e_1+\cdots+e_n[/math]. No degree is associated with the 0 monomial.A polynomial is a sum of finitely many monomials. The degree of a polynomial is defined to be the maximum of the degrees of the monomials that sum to give the polynomial.For instance, the polynomial in three variables [math]x[/math], [math]y[/math], [math]z[/math] given by[math]p(x,y,z) = 2x^3y+3z^2-xyz-3xz[/math]has degree [math]\max\{4,2,3,2\}=4[/math]. [math]\blacksquare[/math]

Consider the equation to be :Ax^3+ Bx^2+CX+D>=0Try to factorize the above equationInitially,find the root of the equation which can be computed by hit and try methodTry the different factors of D in the equation and check for which factor the equation is 0,Once  you've got that you can easily factorize  by long division method or by another easier methodI don't no what it is called but it's quick,Suppose z is a value that satisfies the equationWe can factorize it by,Ax^2(x-z)+Ex(x-z)+F(x-z),Where E and F are constants,=>(x-z)(Ax^2+Ex+F)Now Factorize 2 degree polynomialBy splitting the middle term,To get the result simply: Apply  the condition for each factor(x-a)(x-b)(x-c)>0Thenx>a and x>b and x>c.

One method, which some teachers consider to be a form of cheating, is to put the polynomial into a graphing calculator or a graphing website and observe approximately where the graph crosses the x-axis.You can then zoom in on those locations to get a better estimate, or use the calculator’s function to find a zero of that equation. (You may have to do it several times).Here is an example with an equation I made up for purposes of a demonstration.Equation: [math]x^7 - 8x^6 - 125x^5 +664x^4 + 4684x^3 - 15632x^2 - 47040x + 112896 = 0[/math]This one has easy solutions, but you will frequently find that your answers are irrational, so you’ll have to zoom in to get closer to an answer.One of the reasons I don’t mind doing this for my homework is that it helps me figure out how many answers I am looking for.Can you imagine trying to factor such an equation in the days of the slide rule? When I first took Algebra, Geometry, Calculus and Trigonometry, we had to use paper and pencil, with tables printed in the back of our books. Simple calculators were not yet on the market in my high school years.

There was a whole subject devoted to the problem of solving polynomial equations called the theory of equations.  It was sometimes taught as a college course in the first half of the 20th century. The problem of determining whether there were any roots of polynomials of even degree ≥ 4 was one of the difficult parts of the course.  Sometimes they have roots, sometimes they don't.  Here's one that just misses having roots:Odd degree equations are no problem.  You can find a root of an odd polynomial with ease since for large negative values their signs are opposite of their signs for large positive values, so by the intermediate value theorem they're 0 somewhere.  You can use the bisection method, Horner's method, or Newton's method to find a root.But what about 6th degree and other even degree polynomials?  Their derivatives are lower degree equations, and you'll be able to find all their roots recursively.  That means you know the maxima and minima of the original polynomial, so you can find its roots.The usual method that's used is something different, however.  It's called Sturm's theorem.  You can find it described on page 86 of Leonard Eugene Dickson's First Course in the Theory of Equations, Wiley & Sons, 1922.  See the Project Gutenburg edition on line at gutenberg.org.

You don’t. The characteristic polynomial doesn’t tell you what the degree of the minimal polynomial is.Here are two matrices:[math]A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}[/math][math]B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}[/math]The characteristic polynomial of both matrices is the same: [math]p_A(X)=p_B(X)=(X-1)^2[/math]. However, the minimal polynomial of [math]A[/math] is [math]m_A(X) = X-1[/math] (degree [math]1[/math]) while the minimal polynomial of [math]B[/math] is [math]m_B(X)=p_B(X)=(X-1)^2[/math] (degree [math]2[/math]).You do know that the minimal polynomial must divide the characteristic polynomial, which is often a very useful first step in finding it. But you have to work with the actual matrix to find the degree of the minimal polynomial. Knowing the characteristic polynomial isn’t sufficient.There are various approaches and methods for doing that, and it’s hard to offer specific advice without context.