Question

81. Find a cubic polynomial whose zeroes are 3, and -1.
2​

Answer 1

$\sf\blue{Correct \ Question}$

$\sf{Find \ a \ cubic \ polynomial \ whose \ zeroes}$

$\sf{are \ 3, \ -1 \ and \ 2.}$

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$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ required \ cubic \ polynomial \ is}$

$\sf{x^{3}-4x^{2}+x+6.}$

$\sf\orange{Given:}$

$\sf{\implies{Zeroes \ of \ the \ polynomial \ are}}$

$\sf{3, \ -1 \ and \ 2.}$

$\sf\pink{To \ find:}$

$\sf{Cubic \ polynomial \ whose \ zeroes \ are}$

$\sf{3, \ -1 \ and \ 2.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ \alpha \ be \ 3, \ \beta \ be \ -1 \ and \ \gamma \ be \ 2.}$

$\sf{Sum \ of \ zeroes=\alpha+\beta+\gamma}$

$\sf{=3+(-1)+2}$

$\sf{\therefore{Sum \ of \ zeroes=4...(1)}}$

$\sf{Sum \ of \ product \ of \ zeroes=\alpha\beta+\alpha\gamma+\beta\gamma}$

$\sf{=3(-1)+3(2)+2(-1)}$

$\sf{=-3+6-2}$

$\sf{\therefore{Sum \ of \ product \ of \ zeroes=1...(2)}}$

$\sf{Product \ of \ zeroes=\alpha\beta\gamma}$

$\sf{=3(-1)(2)}$

$\sf{\therefore{Product \ of \ zeroes=-6...(3)}}$

$\sf{Cubic \ polynomial}$

$\sf{\implies{x^{3}-(\alpha+\beta+\gamma)x^{2}+(\alpha\beta+\alpha\gamma+\beta\gamma)x-(\alpha\beta\gamma)}}$

$\sf{...from \ (1), \ (2) \ and \ 3}$

$\sf{\implies{x^{3}-4x^{2}+x+6}}$

$\sf\purple{\tt{\therefore{The \ required \ cubic \ polynomial \ is}}}$

$\sf\purple{\tt{x^{3}-4x^{2}+x+6.}}$