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| 1. |
Verify that 1/2 1 - 2 are zeroes of cubic polynomial |
| Answer» {tex}2 \\mathrm { x } ^ { 3 } + \\mathrm { x } ^ { 2 } - 5 \\mathrm { x } + 2 ; \\frac { 1 } { 2 } , 1 , - 2{/tex}Comparing the given polynomial with\xa0{tex}a x ^ { 3 } + b x ^ { 2 } + c x + d{/tex}, we geta = 2, b = 1, c = 5, d = 2Let\xa0{tex}p ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}Then,{tex}P \\left( \\frac { 1 } { 2 } \\right) = 2 \\left( \\frac { 1 } { 2 } \\right) ^ { 3 } + \\left( \\frac { 1 } { 2 } \\right) ^ { 2 } - 5 \\left( \\frac { 1 } { 2 } \\right) + 2{/tex}{tex}= \\frac { 1 } { 4 } + \\frac { 1 } { 4 } - \\frac { 5 } { 2 } + 2 = 0{/tex}{tex}p ( 1 ) = 2 ( 1 ) ^ { 3 } + ( 1 ) ^ { 2 } - 5 ( 1 ) + 2{/tex}= 2 + 1 - 5 + 2 = 0{tex}p ( - 2 ) = 2 ( - 2 ) ^ { 3 } + ( - 2 ) ^ { 2 } - 5 ( - 2 ) + 2{/tex}= -16 + 4 + 10 + 2 = 0Therefore,\xa0{tex}\\frac{1}{2}{/tex}, 1 and -2 are the zeroes of{tex}2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}So,\xa0{tex}\\alpha = \\frac { 1 } { 2 } , \\beta = 1 \\text { and } \\gamma = - 2{/tex}Therefore,{tex}\\alpha + \\beta + \\gamma = \\frac { 1 } { 2 } + 1 + ( - 2 ) = - \\frac { 1 } { 2 } = - \\frac { b } { a }{/tex}{tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = \\left( \\frac { 1 } { 2 } \\right) \\times ( 1 ) + ( 1 ) \\times ( - 2 ) + ( - 2 ) \\times \\left( \\frac { 1 } { 2 } \\right){/tex}{tex}= \\frac { 1 } { 2 } - 2 - 1 = - \\frac { 5 } { 2 } = \\frac { c } { a }{/tex}{tex}\\alpha \\beta \\gamma = \\left( \\frac { 1 } { 2 } \\right) \\times ( 1 ) \\times ( - 2 ) = - 1 = \\frac { - 2 } { 2 } = \\frac { - d } { a }{/tex} | |