Question

What is the general form of a cubic polynomial in terms of it's roots?(like quadratic polynomials are xsquare-sx+p)

Answer 1

Solutions Of Equations And Interpolation by K.A. Stroud (Advanced Engineering Math) page 4. This is how it goes

Let αα,ββ,γγ be the roots of x3+px2+qx+r=0x3+px2+qx+r=0. Then,writing the expression x3+px2+qx+rx3+px2+qx+rin terms of αα, ββ and γγ gives (x−α)(x−β)(x−γ)(x−α)(x−β)(x−γ).

∴∴ x3+px2+qx+r=(x−α)(x−β)(x−γ)x3+px2+qx+r=(x−α)(x−β)(x−γ).

=(x2−[α+β]x+αβ)(x−γ)=(x2−[α+β]x+αβ)(x−γ)

=x3−(α+β)x2+αβx−γx2+(α+β)γx−αβγ=x3−(α+β)x2+αβx−γx2+(α+β)γx−αβγ

=x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ=x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ

∴∴ equating coefficients

(a) α+β+γ=−pα+β+γ=−p.

(b) αβ+βγ+γα=qαβ+βγ+γα=q.

(c) αβγ=−rαβγ=−r.

This, of course, applies to a cubic equation. Let us extend this to a more general equation

In general, if α1α1, α2α2, α3,…,αnα3,…,αn are the roots of the equation p0xn+p1xn−1+p2xn−2+⋯+pn−1x+pn=0p0xn+p1xn−1+p2xn−2+⋯+pn−1x+pn=0where (p0≠0p0≠0) then

sum of the roots =−p1p0=−p1p0

sum of products of the roots, two at a time =p2p0=p2p0

sum of products if the roots, three at a time =−p3p0=−p3p0

sum of products of the roots, nn at a time = (−1)n,pnp0(−1)n,pnp0

I was able to understand the cubic equation's part but I am completely lost with the general part (i.e an nnth degree polynomial). I am looking for a simpler explanation of what that means. However I understand that it can be used as formulas for finding the roots but i need to know how did he obtain the above formulas.

Answer 2

x³ + (α+β+γ)x²+(αβ+βγ+αγ)x+αβγ

this is the standard form for cubic polynomial.