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| 1. |
find all the zeroes of 2x^4-3x^3-3x^2+6x-2, if you know that two of its zeroes are √2 and -√2. |
| Answer» Let\xa0{tex}p ( x ) = 2 x ^ { 4 } - 3 x ^ { 3 } - 3 x ^ { 2 } + 6 x - 2{/tex}Here\xa0{tex}\\sqrt 2 {/tex}\xa0and\xa0{tex} - \\sqrt 2 {/tex}\xa0are zeroes of p(x){tex}( x - \\sqrt { 2 } ) ( x + \\sqrt { 2 } ) = \\left( x ^ { 2 } - 2 \\right){/tex}\xa0is a factor of p(x){tex}q ( x ) = 2 x ^ { 2 } - 3 x + 1{/tex}{tex}= 2 x ^ { 2 } - 2 x - x + 1{/tex}{tex}= 2 x ( x - 1 ) - 1 ( x - 1 ){/tex}{tex}= ( 2 x - 1 ) ( x - 1 ){/tex}{tex}\\therefore {/tex}\xa0other two zero\'s arex = 1 and\xa0{tex}x = \\frac{1}{2}{/tex}\xa0 | |