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Find the polynomial p(x), with real coefficients, such that p(2) = 5 and p(x) p(y) = p(x) + p(y) + p(xy) - 2 for all real numbers x and y.

 Nov 21, 2019
 #1
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Find the polynomial \(p(x)\), with real coefficients,
such that \(p(2) = 5\) and \(p(x) p(y) = p(x) + p(y) + p(xy) - 2\) for all real numbers \(x\) and \(y\).

 

\(\begin{array}{|lrcll|} \hline & \mathbf{ p(x) p(y)} &=& \mathbf{ p(x) + p(y) + p(xy) - 2 } \\\\ x=1,\ y=2: & p(1) p(2) &=& p(1) + p(2) + p(1*2) - 2 \quad | \quad \mathbf{p(2) = 5} \\ & p(1) *5 &=& p(1) + 5 + 5 - 2 \\ & 5p(1) &=& p(1) + 8 \\ & 4p(1) &=& 8 \\ & p(1) &=& \dfrac{8}{4} \\ & \mathbf{p(1)} &=& \mathbf{2} \\\\ x=0,\ y=2: & p(0) p(2) &=& p(0) + p(2) + p(0*2) - 2 \quad | \quad \mathbf{p(2) = 5} \\ & p(0) *5 &=& p(0) + 5 + p(0) - 2 \\ & 3p(0) &=& 3 \quad | \quad :3 \\ & \mathbf{p(0)} &=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline & p(0) &=& 1 \\ & p(1) &=& 2 \\ & p(2) &=& 5 \\ \hline & p(x) &=& ax^2+bx+c \\ p(0)=1: & 1 &=& a*0^2+b*0 +c \\ & \mathbf{c} &=& \mathbf{1} \\\\ p(1)=2: & 2 &=& a*1^2+b*1 +c \quad | \quad c=1 \\ & 2 &=& a+b+1 \\ & \mathbf{a+b} &=& \mathbf{1} \\\\ & \mathbf{ b} &=& \mathbf{1-a} \\\\ p(2)=5: & 5 &=& a*2^2+b*2 +c \\ & 5 &=& 4a+2b +c \quad | \quad c=1 \\ & 5 &=& 4a+2b + 1 \\ & 4 &=& 4a+2b \quad | \quad :2 \\ & 2a+b &=& 2 \quad | \quad b=1-a \\ & 2a+1-a &=& 2 \\ & \mathbf{ a} &=& \mathbf{1 } \\\\ & \mathbf{ b} &=& \mathbf{1-a} \quad | \quad a = 1 \\ & b &=& 1-1 \\ & \mathbf{b} &=& \mathbf{0} \\ \hline & p(x) &=& ax^2+bx+c \quad | \quad a=1,\ b=0,\ c=1 \\ & \mathbf{p(x)} &=& \mathbf{x^2+1} \\ \hline \end{array}\)

 

check:

\(\begin{array}{|rcll|} \hline \mathbf{ p(x) p(y)} &=& \mathbf{ p(x) + p(y) + p(xy) - 2 } \\\\ (x^2+1)(y^2+1) &=& (x^2+1) + (y^2+1) + \left((xy)^2+1\right) - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2+1 + y^2+1 + x^2y^2+1 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 1 +1 +1 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 3 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 1\ \checkmark \\ \hline \end{array}\)

 

laugh

 Nov 21, 2019

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