Simplifying a symbolic expression

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MILLER BIGGELAAR 2022-9-25

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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?

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Dyuman Joshi 2022-9-25

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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 =

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Answer 1

Walter Roberson 2022-9-25

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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)

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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 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.

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