polynomials

IIT JEE 2015 Help › Algebra questions › polynomials

p(x) = x⁵ + x² +1 have roots

x₁ , x₂,x₃,x₄,x₅.

g(x) = x²-2. then the value of

g(x₁).g(x₂).g(x₃).g(x₄).g(x₅) - 30g(x₁.x₂.x₃.x₄.x₅) is ?

#Mathematics #Algebra

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6 Answers

1708

man111 singh ·Mar 18 '12 at 7:41

$\hspace{-16}$Let $\mathbf{g(x)=y=x^2-2\Leftrightarrow x^2=(y+2)}$\\\\ Now Given That $\mathbf{x^5+x^2+1=0\Leftrightarrow (x^5)=-(x^2+1)}$\\\\ $\mathbf{x^{10}=x^4+1+2x^2}$\\\\ $\mathbf{(y+2)^5=(y+2)^2+1+2y+4}$\\\\ $\mathbf{y^5+10y^4+40y^3+79y^2+74y+23=0}$\\\\ So $\mathbf{y_{1}.y_{2}.y_{3}.y_{4}.y_{5}=-23}$\\\\ Where $\mathbf{y_{1}\;,y_{2}\;,y_{3}\;,y_{4}\;,y_{5}}$ are the Roots of Given Eq.\\\\ and $\mathbf{x^5+x^2+1=0}$\\\\ So $\mathbf{x_{1}.x_{2}.x_{3}.x_{4}.x_{5}=-1}$\\\\ So $\mathbf{g(x_{1}.x_{2}.x_{3}.x_{4}.x_{5})=(x_{1}.x_{2}.x_{3}.x_{4}.x_{5})^2-2}$\\\\ So $\mathbf{y_{1}.y_{2}.y_{3}.y_{4}.y_{5}-30.(1-2)=-23+30=7}$\\\\ Where $\mathbf{g(x_{i})=y_{i}\forall i=1,2,3,4,5}$$

May be some other beautiful Method, So Rishab if you have so plz post here.

Thanks

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rishabh ·Mar 18 '12 at 9:51

super. nope my method was longer so i posted here :P

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341

Hari Shankar ·Mar 19 '12 at 1:06

$\because \ g(x) = (x-\sqrt 2)(x+\sqrt 2)$ and $x 1 x 2 x 3 x 4 x 5 = - 1$

the given expression equals

$g (\sqrt 2 ) g (- \sqrt 2 ) - 30 g (- 1) = 7$

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rishabh ·Mar 19 '12 at 2:49

@hsbhat sir i guess there's a typo in your post. g(âˆš2) = 0 ...
btw, how do you get anything of that sort directly?

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341

Hari Shankar ·Mar 20 '12 at 4:54

sorry, i meant $p (\sqrt 2 ) p (- \sqrt 2 ) - 30 p (- 1) = 7$

I use $p (x) = (x - x 1 ) (x - x 2 ) (x - x 3 ) (x - x 4 ) (x - x 5 )$

So

$(x 1 - \sqrt 2 ) (x 2 - \sqrt 2 )) (x 3 - \sqrt 2 ) (x 4 - \sqrt 2 ) (x 5 - \sqrt 2 ) = - p (\sqrt 2 )$

Now you can finish

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rishabh ·Mar 20 '12 at 5:38

amazing!

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