How do you write a polynomial in standard form given zeros 1 (multiplicity 2), -2 (multiplicity 3)?

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Answer 1 · 2016-05-26T09:35:22

$x^5+4x^4+x^3-10x^2-4x+8=0$

If ${alpha,beta,gamma,delta,..}$ are the zeros of a function, then the function is

$(x-alpha)(x-beta)(x-gamma)(x-delta)...=0$

Here zeros are $1$ (multiplicity $2$ ) and $-2$ (multiplicity $3$ ), hence function is

$(x-1)(x-1)(x+2)(x+2)(x+2)=0$ or

$(x-1)^2(x+2)^3=0$ or

$(x^2-2x+1)(x^3+6x^2+12x+8)=0$ or

$x^2(x^3+6x^2+12x+8)-2x(x^3+6x^2+12x+8)+1(x^3+6x^2+12x+8)=0$

$x^5+6x^4+12x^3+8x^2-2x^4-12x^3-24x^2-16x+x^3+6x^2+12x+8=0$

$x^5+4x^4+x^3-10x^2-4x+8=0$