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Question

# Suppose α,β,γ are roots of x3+x2+2x+3=0. If f(x)=0 is a cubic polynomial equation whose roots are α+β,β+γ,γ+α then f(x)=

A
x3+2x23x1
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B
x3+2x23x+1
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C
x3+2x2+3x1
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D
x3+2x2+3x+1
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Solution

## The correct option is C x3+2x2+3x+1 Roots of the given Equation x3+x2+2x+3=0 are α,β,γNow ,α+β+γ=−1∑αβ=2αβγ=−3Now, defining a equation having roots α+β,β+γ,γ+α as x3+ax2+bx+c=0So,a=sum of roots=2(α+β+γ)a=−2b = Sum of the roots two at a time = (α+β)(γ+α)+(γ+α)(β+γ)+(β+γ)(α+β)b=α2+β2+γ2+3(αβ+γα+βγ)b=(α+β+γ)2+(αβ+γα+γβ) b=(−1)2+2b=3c = Product of roots = (α+β)(α+γ)(γ+β)c=((α+β+γ)−γ))((α+β+γ)−β)((α+β+γ)−α)this expression is similar to the equation(x−α)(x−β)(x−γ) as x=(α+β+γ)=−1hence, c=(−1)3+1.(−1)2+2.(−1)+3c=1Therefore the required equation is x3+2x2+3x+1=0

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