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Question Number 39879 by Rio Mike last updated on 12/Jul/18

let p(x)= x^3 + 3x − 4 find α+β and αβ

$${let}\: \\ $$$${p}\left({x}\right)=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}\:−\:\mathrm{4} \\ $$$${find}\:\alpha+\beta\:{and}\: \\ $$$$\alpha\beta \\ $$$$ \\ $$$$ \\ $$

Answered by MrW3 last updated on 12/Jul/18

p(x)= x^3 + 3x − 4=(x−1)(x^2 +Ax+B) =x^3 +(A−1)x^2 +(B−A)x−B ⇒B=4 ⇒A−1=0⇒A=1 ⇒B−A=4−1=3=^! 3 p(x)= x^3 + 3x − 4=(x−1)(x^2 +x+4) =(x−1)(x−α)(x−β) ⇒α+β=−1 ⇒αβ=4 is this what you mean?

$${p}\left({x}\right)=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}\:−\:\mathrm{4}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{Ax}+{B}\right) \\ $$$$={x}^{\mathrm{3}} +\left({A}−\mathrm{1}\right){x}^{\mathrm{2}} +\left({B}−{A}\right){x}−{B} \\ $$$$\Rightarrow{B}=\mathrm{4} \\ $$$$\Rightarrow{A}−\mathrm{1}=\mathrm{0}\Rightarrow{A}=\mathrm{1} \\ $$$$\Rightarrow{B}−{A}=\mathrm{4}−\mathrm{1}=\mathrm{3}\overset{!} {=}\mathrm{3} \\ $$$$ \\ $$$${p}\left({x}\right)=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}\:−\:\mathrm{4}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}−\alpha\right)\left({x}−\beta\right) \\ $$$$\Rightarrow\alpha+\beta=−\mathrm{1} \\ $$$$\Rightarrow\alpha\beta=\mathrm{4} \\ $$$${is}\:{this}\:{what}\:{you}\:{mean}? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

x^3 +3x−3−1 x^3 −1+3(x−1) (x−1)(x^2 +x+1)+3(x−1) (x−1)(x^2 +x+4) one root is 1 another two α and β α+β=−1 ∝β=4

$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}−\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}+\mathrm{3}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\mathrm{3}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right) \\ $$$${one}\:{root}\:{is}\:\mathrm{1} \\ $$$${another}\:{two}\:\alpha\:{and}\:\beta \\ $$$$\alpha+\beta=−\mathrm{1}\:\:\:\propto\beta=\mathrm{4} \\ $$

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