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Question Number 58092 by smiak8742 last updated on 17/Apr/19

6x^3 +5x^2 −6x−5=0

$$\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{5}=\mathrm{0} \\ $$

Answered by tanmay last updated on 17/Apr/19

6x^3 −6x+5x^2 −5=0  6x(x^2 −1)+5(x^2 −1)=0  (x^2 −1)(6x+5)=0  (x+1)(x−1)(6x+5)=0  x=−1 ,1 and ((−5)/6)

$$\mathrm{6}{x}^{\mathrm{3}} −\mathrm{6}{x}+\mathrm{5}{x}^{\mathrm{2}} −\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{6}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1}\:,\mathrm{1}\:{and}\:\frac{−\mathrm{5}}{\mathrm{6}} \\ $$

Answered by Kunal12588 last updated on 17/Apr/19

6x^3 +5x^2 −6x−5=0  ⇒x^2 (6x+5)−1(6x+5)=0  ⇒(x^2 −1)(6x+5)=0  ⇒(x−1)(x+1)(6x+5)=0  ⇒x=1 or x =−1 or x=−(5/6)

$$\mathrm{6}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{6}{x}+\mathrm{5}\right)−\mathrm{1}\left(\mathrm{6}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left(\mathrm{6}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\:{or}\:{x}\:=−\mathrm{1}\:{or}\:{x}=−\frac{\mathrm{5}}{\mathrm{6}} \\ $$

Answered by MJS last updated on 17/Apr/19

first try all factors of 5:  x=−5 ⇒ no solution  x=−1 ⇒ solution x_1 =−1  x=1 ⇒ solution x_2 =1  x=5 ⇒ no solution  6(x+1)(x−1)(x−x_3 )=...−5  6×1×(−1)×(−x_3 )=−5  6x_3 =−5  x_3 =−(5/6)

$$\mathrm{first}\:\mathrm{try}\:\mathrm{all}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{5}: \\ $$$${x}=−\mathrm{5}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$${x}=−\mathrm{1}\:\Rightarrow\:\mathrm{solution}\:{x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\mathrm{solution}\:{x}_{\mathrm{2}} =\mathrm{1} \\ $$$${x}=\mathrm{5}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{6}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}−{x}_{\mathrm{3}} \right)=...−\mathrm{5} \\ $$$$\mathrm{6}×\mathrm{1}×\left(−\mathrm{1}\right)×\left(−{x}_{\mathrm{3}} \right)=−\mathrm{5} \\ $$$$\mathrm{6}{x}_{\mathrm{3}} =−\mathrm{5} \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{5}}{\mathrm{6}} \\ $$

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