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Find the zeros of polynomial x3 - 5x2 - 2x+24 if it is given that the product of its two zeros is12

Answer» Let {tex} \\alpha , \\beta , \\gamma{/tex}\xa0be the zeroes of polynomial f(x) = x3 - 5x2 - 2x + 24 such that {tex} \\alpha\\beta{/tex}\xa0= 12.{tex} \\alpha + \\beta + \\gamma = - \\left( - \\frac { 5 } { 1 } \\right) = 5{/tex} ................ (i){tex} \\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = \\frac { - 2 } { 1 } = - 2{/tex}and,\xa0{tex} \\alpha \\beta \\gamma = - \\frac { 24 } { 1 } = - 24{/tex}Putting,\xa0{tex} \\alpha \\beta = 12{/tex}\xa0in\xa0{tex} \\alpha \\beta \\gamma = - 24{/tex}, we get{tex} 12 \\gamma = - 24{/tex}{tex} \\Rightarrow \\gamma = - \\frac { 24 } { 12 } = - 2{/tex}Putting\xa0{tex} \\gamma= -2{/tex} in eq.(i), we get{tex} \\alpha + \\beta - 2 = 5{/tex}{tex} \\Rightarrow \\quad \\alpha + \\beta = 7{/tex}Now,\xa0{tex} ( \\alpha - \\beta ) ^ { 2 } = ( \\alpha + \\beta ) ^ { 2 } - 4 \\alpha \\beta{/tex}{tex} \\Rightarrow \\quad ( \\alpha - \\beta ) ^ { 2 } = 7 ^ { 2 } - 4 \\times 12{/tex}{tex} \\Rightarrow \\quad ( \\alpha - \\beta ) ^ { 2 } = 1{/tex}{tex} \\Rightarrow \\quad \\alpha - \\beta = \\pm 1{/tex}Thus, we have{tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta = 1 {/tex} or, {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex}CASE I: When {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= 1{/tex}Solving {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= 1{/tex} , we get{tex} \\alpha = 4{/tex} and {tex}\\beta= 3{/tex}\xa0CASE II: When {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex}Solving {tex} \\alpha + \\beta= 7{/tex} and {tex} \\alpha - \\beta= -1{/tex} , we get{tex} \\alpha = 3{/tex} and {tex}\\beta= 4{/tex}\xa0.Hence, the zeros of the given polynomial are 3, 4 and - 2 or 4,3 , -2.


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