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Factorise I) 1+x4+x8 II) x4  + 4

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Factorise I) 1+x4+x8

II) x + 4

  • 2 answers

Rashmi Bajpayee 6 years, 8 months ago

||) {tex}{x^4} + 4 = {x^2} + {2^2} = {\left( {x + 2} \right)^2} - 2 \times x \times 2{/tex}         [Since {tex}{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab{/tex}]

{tex}{\left( {x + 2} \right)^2} - 4x{/tex}

{tex}{\left( {x + 2} \right)^2} - {\left( {2\sqrt x } \right)^2}{/tex}

{tex}\left( {x + 2 - 2\sqrt x } \right)\left( {x + 2 + 2\sqrt x } \right){/tex}

{tex}\left( {x + 2\sqrt x + 2} \right)\left( {x - 2\sqrt x + 2} \right){/tex}

 

 

Rashmi Bajpayee 6 years, 8 months ago

|) 1+x<font size="2">4</font>+x<font size="2">8</font><font size="2"> = x8 + x4 + 1</font>

Adding 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}

{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}

{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}

{tex}\left[ {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \right]\left( {{x^4} - {x^2} + 1} \right){/tex}

{tex}\left( {{x^2} + 1 + x} \right)\left( {{x^2} + 1 - x} \right)\left( {{x^4} - {x^2} + 1} \right){/tex}

{tex}\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)\left( {{x^4} - {x^2} + 1} \right){/tex}

 

 

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