Algebra (Olympiad)hardFree

Sum of Two Squares: Find P(1)P(2) for a Quartic Product | IOQM

JEE Maths question with a full step-by-step solution.

Question
Consider the polynomial
(x2+1)(x2+4)(x2−2x+2)(x2+2x+2).(x^2+1)(x^2+4)(x^2-2x+2)(x^2+2x+2).
 Let f(x)=(x4+2x2+2x)2+(P(x))2f(x)=\big(x^4+2x^2+2x\big)^2+\big(P(x)\big)^2, where P(x)P(x) is a cubic polynomial. Find the value of P(1) P(2)P(1)\,P(2)
Solution
Answer: 20
Step 1: Group the four quadratic factors into two pairs:
[(x2+1)(x2+4)][(x2−2x+2)(x2+2x+2)].\big[(x^2+1)(x^2+4)\big]\big[(x^2-2x+2)(x^2+2x+2)\big].
 Step 2: Simplify each pair and write it as a sum of two squares. For the first pair,
(x2+1)(x2+4)=x4+5x2+4=(x2+2)2+x2.(x^2+1)(x^2+4)=x^4+5x^2+4=(x^2+2)^2+x^2.
 For the second pair, group as [(x2+2)−2x][(x2+2)+2x]\big[(x^2+2)-2x\big]\big[(x^2+2)+2x\big]:
(x2−2x+2)(x2+2x+2)=(x2+2)2−(2x)2=x4+4=(x2)2+22.(x^2-2x+2)(x^2+2x+2)=(x^2+2)^2-(2x)^2=x^4+4=(x^2)^2+2^2.
 Step 3: Each factor is now a sum of two squares, so apply the Brahmagupta-Fibonacci identity
(a2+b2)(c2+d2)=(ac+bd)2+(ad−bc)2(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2
 with a=x2+2, b=x, c=x2, d=2a=x^2+2,\ b=x,\ c=x^2,\ d=2:
ac+bd=(x2+2)x2+x⋅2=x4+2x2+2x,ac+bd=(x^2+2)x^2+x\cdot2=x^4+2x^2+2x,
ad−bc=(x2+2)⋅2−x⋅x2=−x3+2x2+4.ad-bc=(x^2+2)\cdot2-x\cdot x^2=-x^3+2x^2+4.
 Therefore
(x2+1)(x2+4)(x2−2x+2)(x2+2x+2)=(x4+2x2+2x)2+(−x3+2x2+4)2.(x^2+1)(x^2+4)(x^2-2x+2)(x^2+2x+2)=\big(x^4+2x^2+2x\big)^2+\big(-x^3+2x^2+4\big)^2.
 Step 4: Comparing with f(x)=(x4+2x2+2x)2+(P(x))2f(x)=\big(x^4+2x^2+2x\big)^2+\big(P(x)\big)^2, the cubic polynomial is
P(x)=−x3+2x2+4.P(x)=-x^3+2x^2+4.
 Step 5: Evaluate at x=1x=1 and x=2x=2:
P(1)=−1+2+4=5,P(2)=−8+8+4=4.P(1)=-1+2+4=5,\qquad P(2)=-8+8+4=4.
 Hence
P(1) P(2)=5×4=20.P(1)\,P(2)=5\times4=20.
 Answer: 20
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