Factorise : 1) 2√2x3+5√2x2-7√2     …

1) 2√2x^{<font size="2">3</font>}+5√2x^{<font size="2">2</font>}-7√2

{tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

Putting x = 1, in p(x),

{tex}p\left( 1 \right) = 2\surd 2{\left( 1 \right)^3} + 5\surd 2{\left( 1 \right)^2} - 7\surd 2{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( 1 \right) = 2\surd 2 + 5\surd 2 - 7\surd 2{/tex} = {tex}7\surd 2 - 7\surd 2 = 0{/tex}

Since p(1) = 0, therefore, (x - 1) is a factor of p(x)

Dividing p(x) by (x -1), we get the quotient {tex}2\sqrt 2 {x^2} + 7\sqrt 2 {/tex}

Therefore,

{tex}p\left( x \right) = 2\surd 2{x^3} + 5\surd 2{x^2} - 7\surd 2{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {2\sqrt 2 {x^2} + 7\sqrt 2 } \right]{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \left( {x - 1} \right)\left[ {\sqrt 2 \left( {2{x^2} + 7} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}     {tex}p\left( x \right) = \sqrt 2 \left( {x - 1} \right)\left( {2{x^2} + 7} \right){/tex}

3) {tex}{p^4} - 81{q^4}{/tex}

= {tex}{\left( {{p^2}} \right)^2} - {\left( {9{q^2}} \right)^2}{/tex}                               [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{p^2} - 9{q^2}} \right]{/tex}

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {{{\left( p \right)}^2} - {{\left( {3q} \right)}^2}} \right]{/tex}

= {tex}\left[ {{p^2} + 9{q^2}} \right]\left[ {\left( {p + 3q} \right)\left( {p - 3q} \right)} \right]{/tex}        [Using identity {tex}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right){/tex}]

= {tex}\left( {{p^2} + 9{q^2}} \right)\left( {p + 3q} \right)\left( {p - 3q} \right){/tex}

2) p(x) = x^{<font size="2">3</font>}+13x^{<font size="2">2</font>}+32x+20

putting x = -1 in given polynomial,

p(-1) = (-1)^{<font size="2">3</font>}+ 13(-1)^{<font size="2">2</font>}+ 32(-1) +20

         = -1 + 13 - 32 + 20 = 0

Since p(-1) = 0, therefore (x + 1) is a factor of p(x).

Now, on dividing p(x) by (x + 1), we get the quotient x² + 12x + 20

{tex}p\left( x \right) = {x^3} + 13{x^2} + 32x + 20{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 12x + 20} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {{x^2} + 10x + 2x + 20} \right]{/tex}

{tex}\Rightarrow{/tex}          {tex}p\left( x \right) = \left( {x + 1} \right)\left[ {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right]{/tex}

{tex}\Rightarrow{/tex}         {tex}p\left( x \right) = \left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 10} \right){/tex}