Find cubic polynomial whose zeroes are 2,1,1

1.	Find cubic polynomial whose zeroes are 2,1,1
Answer» Let\xa0{tex}\\alpha,\\mathrm\\beta\\;\\mathrm{and}\\;\\mathrm\\gamma{/tex}\xa0be the zeroes of the given polynomial.\xa0Then, we have\xa0{tex}\\alpha{/tex}\xa0= 2,\xa0{tex}\\beta{/tex}\xa0= 1\xa0and\xa0{tex}\\gamma{/tex}\xa0= 1\xa0Hence{tex}\\alpha + \\beta + \\gamma{/tex}\xa0= 2\xa0+ 1\xa0+\xa01\xa0= 4 ...............(1){tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha{/tex}\xa0= 2(1) + 1(1) + 1(2)\xa0= 2\xa0+ 1\xa0+ 2\xa0= 5 ................(2){tex}\\alpha \\beta \\gamma{/tex}\xa0= 2(1)(1) = 2 .............(3)Now, a cubic polynomial whose zeros are {tex}\\alpha , \\beta{/tex}\xa0and\xa0{tex}\\mathrm\\gamma{/tex}\xa0is equal top(x) = x3\xa0-\xa0{tex}( \\alpha + \\beta + \\gamma ) x ^ { 2 } + ( \\alpha \\beta + \\beta y + \\gamma \\alpha ) x - \\alpha \\beta \\gamma{/tex}On substituting values from (1),(2) and (3) we get{tex}\\mathrm p(\\mathrm x)=\\mathrm x^3-(4)\\mathrm x^2+(5)\\mathrm x-(2){/tex}= x3\xa0- 4x2\xa0+\xa05x - 2

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