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Given that √2 is a zero of cubic polynomial 6x^3+√2x^2-10x-4√2, find its other two zeros |
| Answer» Assume f(x) = {tex}6{x^3} + \\sqrt 2 {x^2} - 10x - 4\\sqrt 2{/tex}If {tex}\\sqrt 2{/tex}\xa0is the zero of f(x), then {tex}(x - \\sqrt 2 ){/tex}\xa0will be a factor of f(x). So, by remainder theorem when f(x) is divided by {tex}(x - \\sqrt 2 ){/tex}, the quotient comes out to be quadratic.Now we divide {tex}6{x^3} + \\sqrt 2 {x^2} - 10x - 4\\sqrt 2{/tex}\xa0by\xa0{tex}(x - \\sqrt 2 ){/tex}.{tex}\\therefore \\;f(x) = (x - \\sqrt 2 )(6{x^2} + 7\\sqrt 2 x + 4){/tex}\xa0(By Euclid’s division algorithm){tex}= (x - \\sqrt 2 )(6{x^2} + 4\\sqrt 2 x + 3\\sqrt 2 x + 4){/tex}\xa0( By factorization method )For zeroes of f(x), put f(x) = 0{tex}\\therefore (x - \\sqrt 2 )(6{x^2} + 4\\sqrt 2 x + 3\\sqrt 2 x + 4) = 0{/tex}{tex}\\Rightarrow \\;(x - \\sqrt 2 )[2x(3x + 2\\sqrt 2 ){/tex}\xa0{tex}+ \\sqrt 2 (3x + 2\\sqrt 2 )] = 0{/tex}{tex}\\Rightarrow (x - \\sqrt 2 )(3x + 2\\sqrt 2 )(2x + \\sqrt 2 ) = 0{/tex}{tex}\\Rightarrow x - \\sqrt 2 = 0{/tex}\xa0or {tex}3x + 2\\sqrt 2 = 0{/tex}\xa0or {tex}2x + \\sqrt 2 = 0{/tex}{tex}\\Rightarrow x = \\sqrt 2{/tex}\xa0or {tex}x = \\frac{{ - 2\\sqrt 2 }}{3}{/tex}\xa0or {tex}x = \\frac { - \\sqrt { 2 } } { 2 }{/tex}So, other two roots are {tex}= \\frac{{ - 2\\sqrt 2 }}{3}{/tex}\xa0and {tex}\\frac{{ - \\sqrt 2 }}{2}{/tex}. | |