Simplifying a symbolic expression

3 次查看（过去 30 天）
MILLER BIGGELAAR
MILLER BIGGELAAR 2022-9-25
评论： Walter Roberson 2022-9-25
Hi team,
I am trying to simplify the symbolic expression below into the form (x + __)(x + __)(x + __)(x + __)(x + __)
syms x k
f = 50*x^5+994*x^4+5504*x^3+20*k*x^3+6233*x^2+170*k*x^2+980*k*x-8732*x+1700*k-24913;
I have tried using the simplify() function however it won't simplify further, approximations are fine. Any suggestions?
  1 个评论
Dyuman Joshi
Dyuman Joshi 2022-9-25
Ran in:
I don't think you will be able to do such factorisation, in an expression where there is an unknown independent variable.
syms f(x,k)
f(x,k) = 50*x^5+994*x^4+5504*x^3+20*k*x^3+6233*x^2+170*k*x^2+980*k*x-8732*x+1700*k-24913;
fac=factor(f)
fac(x, k) = 
Though you will get a factorization for values of k
y=simplify(f(x,0),10)
y = 
z=factor(f(x,0),x,'FactorMode','complex')
z = 

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Walter Roberson
Walter Roberson 2022-9-25
You are trying to find the symbolic roots of a degree 5 polynomial. It does not happen to be one of the quintic polynomials that factors into algebraic numbers (at least not for the general case)
  2 个评论
MILLER BIGGELAAR
MILLER BIGGELAAR 2022-9-25
编辑：MILLER BIGGELAAR 2022-9-25
so there is no way to factorise this into the following then from what i gather?
f*(x+a)(x+b)(x+c)(x+kd)
Even with say, 5+ decimals for each value?
Walter Roberson
Walter Roberson 2022-9-25
You have x^5 and will not be able to recreate that by multiplying four (x+something) terms.
If you had specific numeric k then you could vpasolve() or use sym2poly() and roots(). But with symbolic k there you cannot get a decimal approximation.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by