Verified answer

Answer:

[tex]\boxed{ \sf{ \:\bf \: {x}^{3} - {3x}^{2} - 9x - 5 = \: (x + 1)(x - 5)(x + 1) \: }}\\ \\ [/tex]

Step-by-step explanation:

Given expression is

[tex]\sf \: {x}^{3} - {3x}^{2} - 9x - 5 \\ \\ [/tex]

Let assume that

[tex]\sf \: f(x) = {x}^{3} - {3x}^{2} - 9x - 5 \\ \\ [/tex]

We have to first the one zero of f(x) by using hit and trial method.

Let assume that x = - 1

So, above expression can be rewritten as

[tex]\sf \: f( - 1) = {( - 1)}^{3} - {3( - 1)}^{2} - 9( - 1) - 5 \\ \\ [/tex]

[tex]\sf \: f( - 1) = - 1 - 3 + 9 - 5 \\ \\ [/tex]

[tex]\sf \: f( - 1) = - 4 + 4 \\ \\ [/tex]

[tex]\sf\implies \sf \: f( - 1) = 0 \\ \\ [/tex]

[tex]\sf\implies \sf \: x + 1 \: is \: a \: factor \: of \: f(x) \\ \\ [/tex]

Now, using Long Division Method, we get

[tex]\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - 4x - 5\:\:}}}\\ {\underline{\sf{x + 1}}}& {\sf{\: {x}^{3}  -  {3x}^{2} - 9x - 5 \:\:}} \\{\sf{}}& \underline{\sf{ \: -  {x}^{3} \: - {x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\ {{\sf{}}}& {\sf{ \: \: \: -  4{x}^{2} - 9x - 5}} \\{\sf{}}& \underline{\sf{{4x}^{2} + 4x}} \\ {\underline{\sf{}}}& {\sf{ \qquad \: \: \: \: - 5x - 5}} \\{\sf{}}& \underline{\sf{\qquad \: \: \: \: \: \: 5x + 5}} \\ {\underline{\sf{}}}& {\sf{\qquad \: \: \: \: \: \: \: 0}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \\ [/tex]

So, using Division algorithm method, we get

[tex]\sf \: {x}^{3} - {3x}^{2} - 9x - 5 \\ \\ [/tex]

[tex]\qquad\sf \: = \: (x + 1)( {x}^{2} - 4x - 5) \\ \\ [/tex]

[tex]\qquad\sf \: = \: (x + 1)( {x}^{2} - 5x + x - 5) \\ \\ [/tex]

[tex]\qquad\sf \: = \: (x + 1)[ x(x - 5) + 1(x - 5)]\\ \\ [/tex]

[tex]\qquad\sf \: = \: (x + 1)(x - 5)(x + 1)\\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: {x}^{3} - {3x}^{2} - 9x - 5 = \: (x + 1)(x - 5)(x + 1)\\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

Step-by-step explanation:

Factorise :- x³-3x²-9x-5