Factorise: 1) x4+4x2+16       …

{tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

Putting x = -1 in p(x),

{tex}p\left( { - 1} \right) = {\left( { - 1} \right)^3} - 10{\left( { - 1} \right)^2} - 53\left( { - 1} \right) - 42{/tex} = {tex} - 1 - 10 + 53 - 42 = 0{/tex}

Since {tex}p\left( { - 1} \right) = 0{/tex}, therefore, (x + 1) is a factor of p(x)

Now, on dividing p(x) by (x + 1), we get the quotient {tex}{x^2} - 11x - 42{/tex}

Therefore, {tex}p\left( x \right) = {x^3} - 10{x^2} - 53x - 42{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 11x - 42} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} - 14x + 3x - 42} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x - 14} \right) + 3\left( {x - 14} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 14} \right){/tex}

1) x^{<font size="2">4</font>}+4x^{<font size="2">2</font>}+16

= {tex}{x^4} + 16 + 4{x^2}{/tex}

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

= {tex}{\left( {{x^2} + 4} \right)^2} - 8{x^2} + 4{x^2}{/tex}

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

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

= {tex}\left( {{x^2} + 4 + 2x} \right)\left( {{x^2} + 4 - 2x} \right){/tex}                 [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

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