Question

x^3+2x^2-8x+5=0 , find zeroes

Answer 1

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Answer 2

Hello Friend...

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The answer of u r question is...
${x}^{2} + 2 {x}^{2} - 8x + 5 = 0$
Ans:

Given,

${x}^{2} + 2 {x}^{2} - 8x + 5 = 0$

${2}^{3} - {2}^{2} + {3x}^{2} - 3x - 5x + 5 = 0$

${x}^{2} \times (x - 1) + 3x \times (x - 1 - 5(x - 1) = 0$

$(x - 1) \times ( {2}^{2} + 3x - 5) = 0$

$x = 1 \alpha$

$x \times \frac{ - 3 + \sqrt{2} } {2}$


$x \times \frac{ - 3 - \sqrt{2} }{2}$


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