Question

if the product of two zeroes of the polynomial of f(x) =2x^3+6x^2-4x +9 is then its third zero is​

Answer 1

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if the product of two zeroes of the polynomial of f(x) =2x^3+6x^2-4x +9 is 3 then its third zero is

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$\tt \large \: p(x) = {2x}^{3} + {6x}^{2} - 4x + 9$

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$\tt \star \: let \: \: \alpha , \beta , \gamma \: \: be \: \: the \: \: 3 \: \: roots \: \: \\ \\ \tt \star \alpha \beta = 3$

we know that,

• a=2
• b=6
• c=-4
• d=9

$\tt \large \: \: find \: \: the \: \: products \: \: of \: \: all \: \: zeroes \: \: \\ \bf \: \alpha \beta \gamma = \frac{ - d}{a} = \frac{ - 9}{2}$

$\tt \: \: third \: \: zeroes \: \\ \\ \tt \: given \: \: that \: \: ( \alpha \beta = 6) \\ \\ \sf \large \implies \: 3 \gamma = \frac{ - 9}{2} \\ \\ \sf \large \implies \: \gamma = \frac{ \frac{ - 9}{2} }{3} \\ \\ \sf \large \implies \: \gamma = \frac{ - 9}{3 \times 2} \\ \\ \sf \large \implies \: \gamma = \frac{ - 9}{6} \\ \\ \sf \large \implies \: \gamma = \frac{ - 3}{2}$

Third zero is -3/2