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# Polynomial Problem in Pre-RMO today !

Given a polynomial :

$p(x) = x^5 - 3x^4 + 5x^3 - 2x^2 + 9x - 7 = 0$

With $$\alpha , \beta , \gamma$$ and $$\sigma$$ as its roots .

Find :

$(1 + \alpha^2)(1+\beta^2)(1+\gamma^2)(1+\sigma^2)$

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

Note by Priyansh Sangule
3 years, 9 months ago

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Presumably, you mean the quintic polynomial to have roots $$\alpha,\beta,\gamma,\delta,\epsilon$$, and want to know $(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)(1+\epsilon^2)$ Find the monic quintic polynomial $$g(y)$$ whose roots are $$\alpha^2,\beta^2,\gamma^2,\delta^2,\epsilon^2$$. Do this by substituting $$x=\sqrt{y}$$ and eliminating the square root. Then consider $$g(-1)$$. · 3 years, 9 months ago

Let $$p(x) = x^5-3x^4+5x^3-2x^2+9x-7 = (x-\alpha)\cdot(x-\beta)\cdot(x-\gamma)\cdot (x-\delta)$$

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

$$i^5-3i^4+5i^3-2i^2+9i-7 = (5i-8) = (\alpha - i)\cdot(\beta-i)\cdot(\gamma-i)\cdot(\delta-i)$$

$$-i^5-3i^4-5i^3-2i^2-9i-7 = -(5i-8) = (\alpha + i)\cdot(\beta+i)\cdot(\gamma+i)\cdot(\delta+i)$$

Now multiply these two, we get $$89 = (1+\alpha^2)\cdot(1+\beta^2)\cdot(1+\gamma^2)\cdot(1+\delta^2)$$ · 3 years, 9 months ago