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1 is a zero of 7x-x3-6 polynomial find other zero |
| Answer» The given polynomial is -x3 + 7x - 6 and let\xa0f (x) = -x3 + 7x - 6 .Since 1 is a zero of f(x), so (x - 1) is a factor of f(x).Now we divide f(x) = -x3 + 7x - 6\xa0by (x - 1), we obtain\xa0Where quotient =\xa0(-x2\xa0- x + 6){tex}\\therefore{/tex}\xa0f(x) = (-x3\xa0+ 7x -\xa06) = (x - 1)(-x2\xa0- x + 6){tex}= - ( x - 1 ) \\left( x ^ { 2 } + x - 6 \\right) = - ( x - 1 ) \\left( x ^ { 2 } + 3 x - 2 x - 6 \\right){/tex}{tex}= ( 1 - x ) [ x ( x + 3 ) - 2 ( x + 3 ) ] = ( 1 - x ) ( x + 3 ) ( x - 2 ){/tex}{tex}\\therefore{/tex}\xa0f(x) = 0 {tex}\\Rightarrow{/tex}\xa0{tex}( 1 - x ) ( x + 3 ) ( x - 2 ){/tex} = 0{tex}\\Rightarrow{/tex}\xa0(1 - x ) = 0 or (x + 3) = 0 or (x - 2) = 0{tex}\\Rightarrow{/tex}\xa0x =\xa01 or x = -3 or x = 2.Thus, the other zeros are -3 and 2 | |