Question

find the zeroes of the polynomial x*square-2x-8 and verify the relation between the coefficients and the zeroes of the polynomial

Answer 1

${x}^{2} - 2x - 8$

${x}^{2} - 4x + 2x - 8$

$x(x - 4) + 2(x - 4)$

$(x + 2)(x - 4)$

$\large{verification}$

$x = - 2$

$x = + 4$

$\alpha + \beta = - 2 + 4 = 2 = \frac{ - (coeficientofx}{coeficientofxsquare} = \frac{ - ( - 2)}{1} = 2$

$\alpha \times \beta = - 2 \times 4 = - 8 = \frac{constant \: term}{coeficent \: of \: xsquare} = \frac{ - 8}{1 } = - 8$

$\large{letsrock}$

Answer 2

$\Large{\underline{\underline{\bf{Answer:-}}}}$

${x}^{2} - 2x - 8 \\ \\ {x}^{2} - 2x - 8 = 0 \\ \\ {x}^{2} - 4x + 2x - 8 \\ \\ x(x - 4)2(x - 4) \\ \\ (x + 2)(x - 4) \\ \\ x + 2 = 0 \\ \\ x = - 2 \\ \\ x - 4 = 0 \\ \\ x = 4$

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➣ Let the zeroes of the polynomial be

$\alpha \: and \: \beta$

$sum \: of \: zeroes \: ( \alpha \: and \: \beta ) = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ \\ - 2 + 4 = \frac{2}{1 } \\ \\ 2 = 2$

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$product \: of \: zeroes ( \alpha \beta ) = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ - 2 \times 4 = \frac{ - 8}{1} \\ \\ - 8 = - 8$

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$\Large{\underline{\underline{\bf{Hence\: Verified\: !}}}}$

$\Large{\underline{\underline{\bf{Thanks}}}}$