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Factorise I) 1+x4+x8II) x4\xa0\xa0+ 4 |
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Answer» ||)\xa0{tex}{x^4} + 4 = {x^2} + {2^2} = {\\left( {x + 2} \\right)^2} - 2 \\times x \\times 2{/tex} [Since\xa0{tex}{a^2} + {b^2} = {\\left( {a + b} \\right)^2} - 2ab{/tex}]=\xa0{tex}{\\left( {x + 2} \\right)^2} - 4x{/tex}=\xa0{tex}{\\left( {x + 2} \\right)^2} - {\\left( {2\\sqrt x } \\right)^2}{/tex}=\xa0{tex}\\left( {x + 2 - 2\\sqrt x } \\right)\\left( {x + 2 + 2\\sqrt x } \\right){/tex}=\xa0{tex}\\left( {x + 2\\sqrt x + 2} \\right)\\left( {x - 2\\sqrt x + 2} \\right){/tex}\xa0\xa0 |)\xa01+x4+x8 = x8 + x4 + 1Adding and subtracting 2(x4)(1){tex}{\\left( {{x^4}} \\right)^2} + 2\\left( {{x^4}} \\right)\\left( 1 \\right) + {\\left( 1 \\right)^2} - 2\\left( {{x^4}} \\right) + {x^4} = {\\left( {{x^4} + 1} \\right)^2} - {x^4}{/tex}=\xa0{tex}\\left( {{x^4} + 1 + {x^2}} \\right)\\left( {{x^4} + 1 - {x^2}} \\right) = \\left( {{x^4} + {x^2} + 1} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left[ {{{\\left( {{x^2}} \\right)}^2} + 2\\left( {{x^2}} \\right)\\left( 1 \\right) + 1 - 2\\left( {{x^2}} \\right)\\left( 1 \\right) + {x^2}} \\right]\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left[ {{{\\left( {{x^2} + 1} \\right)}^2} - {x^2}} \\right]\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left( {{x^2} + 1 + x} \\right)\\left( {{x^2} + 1 - x} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left( {{x^2} + x + 1} \\right)\\left( {{x^2} - x + 1} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}\xa0\xa0 |
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