A Quadratic Polynomial System 🤩

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[@JR13751]

You can use first method because polynomials form a field.

[@yoav613]

Nice

[@rakenzarnsworld2]

x^2(x^2-1)

[@Secret_Service]

İts reminding me the yks exam or the known name nightmare of students

[@scottleung9587]

I got the first solution, but not the second one.

[@stephensu4371]

my first impression： （A - B）^2😂

[@golddddus]

You forgot the trivial solution P(x)=0 and Q(x)=0. The third method is the assumption of two trinomial polynomials. This is reached by logical consideration. It is also easy to conclude that the free coefficients of those trinomials are equal to zero. Let P(x)=ax^2+bx and Q(x)=cx^2+dx. We get the equations 2P(x)-Q(x)=x^2+3x, that is, 2a-c=1 and 2b-d=3. From P(x)Q(x)=x^4-x^2 it follows ac=1 and bd=-1. (a1,b1,c1,d1)=(1, 1, 1 ,-1) and (a2,b2,c2,d2)=(-1/2, 1/2, -2, -2). These are the only winning combinations, since some solutions are rejected due to ac+bd=0.😎

[@TheEricthefruitbat]

P=Q=0 is not a solution, because it doesn't satisfy the initial relations - which must hold for all values of x.

[@bkkboy-cm3eb]

2P-Q=x²+3x
PQ=x⁴-x²=x²(x²-1)=x²(x+1)(x-1)
2P(1)-Q(1)=4
P(1)Q(1)=0
→P(1)=2, Q(1)=0 or P(1)=0, Q(1)=-4
2P(0)-Q(0)=0
P(0)Q(0)=0
→P(0)=Q(0)=0
(1) when P(1)=2, Q(1)=0,
→Q(x)=x(x-1)R(x), P(x)=xS(x)
→R(x)S(x)=x+1→ R(x)=1, S(x)=x+1
∴P(x)=x(x+1) , Q(x)=x(x-1)
(2) when P(1)=0, Q(1)=-4,
Q(x)=xR(x), P(x)=x(x-1)S(x)
→R(x)S(x)=x+1
R(1)S(1)=2≠0
∴R(x), S(x) not exist.

[@golddddus]

(→R(x)S(x)=x+1 R(1)S(1)=2≠0 →R(x)S(x)=x+1). It has to be R(-1)S(-1)=0 then P(x)=-0.5*x^2 +0.5*x and Q(x)=-2x^2-2x.😎

[@bkkboy-cm3eb]

@@golddddus
P(x)=0.5x²+0.5x →not satisfied P(1)=0
Q(x)=2x²-2x → not satisfied Q(1)=-4