Polynomial Problem in Pre-RMO today !

Given a polynomial :

\[ p(x) = x^5 - 3x^4 + 5x^3 - 2x^2 + 9x - 7 = 0 \]

With α,β,γ \alpha , \beta , \gamma and σ \sigma as its roots .

Find :

(1+α2)(1+β2)(1+γ2)(1+σ2) (1 + \alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2)

Please help .

(Not pretty sure about some coefficients, but the idea is same.)

Note by Priyansh Sangule
7 years, 8 months ago

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Comments

Presumably, you mean the quintic polynomial to have roots α,β,γ,δ,ϵ\alpha,\beta,\gamma,\delta,\epsilon, and want to know (1+α2)(1+β2)(1+γ2)(1+δ2)(1+ϵ2) (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)(1+\epsilon^2) Find the monic quintic polynomial g(y)g(y) whose roots are α2,β2,γ2,δ2,ϵ2\alpha^2,\beta^2,\gamma^2,\delta^2,\epsilon^2. Do this by substituting x=yx=\sqrt{y} and eliminating the square root. Then consider g(1)g(-1).

Mark Hennings - 7 years, 8 months ago

Let p(x)=x53x4+5x32x2+9x7=(xα)(xβ)(xγ)(xδ)p(x) = x^5-3x^4+5x^3-2x^2+9x-7 = (x-\alpha)\cdot(x-\beta)\cdot(x-\gamma)\cdot (x-\delta)

Now put x=ix = i and x=ix= -i respectively

i53i4+5i32i2+9i7=(5i8)=(αi)(βi)(γi)(δi)i^5-3i^4+5i^3-2i^2+9i-7 = (5i-8) = (\alpha - i)\cdot(\beta-i)\cdot(\gamma-i)\cdot(\delta-i)

i53i45i32i29i7=(5i8)=(α+i)(β+i)(γ+i)(δ+i)-i^5-3i^4-5i^3-2i^2-9i-7 = -(5i-8) = (\alpha + i)\cdot(\beta+i)\cdot(\gamma+i)\cdot(\delta+i)

Now multiply these two, we get 89=(1+α2)(1+β2)(1+γ2)(1+δ2)89 = (1+\alpha^2)\cdot(1+\beta^2)\cdot(1+\gamma^2)\cdot(1+\delta^2)

jagdish singh - 7 years, 8 months ago
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