Answer:

let's factorize the expression \(x^3 - 3x^2 - 9x - 5\) using different methods.

**Method 1: Factor by Grouping**

Step 1: Group the terms in pairs:

\(x^3 - 3x^2 - 9x - 5 = (x^3 - 3x^2) + (-9x - 5)\)

Step 2: Factor out the common factors from each group:

\(x^3 - 3x^2 - 9x - 5 = x^2(x - 3) - 5(3x + 1)\)

Step 3: Factor out the common factor from the whole expression:

\(x^3 - 3x^2 - 9x - 5 = (x^2 - 5)(x - 3)\)

**Method 2: Using Rational Root Theorem**

The rational root theorem states that if \(p/q\) is a rational root of a polynomial with integer coefficients, then \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient.

In this case, the leading coefficient is 1, and the constant term is -5. The possible rational roots are \(\pm 1, \pm 5\).

By testing these roots, we find that \(x = 5\) is a root of the polynomial.

Step 1: Divide the polynomial by \(x - 5\) using long division or synthetic division.

\(x^3 - 3x^2 - 9x - 5 = (x - 5)(x^2 + 2x + 1)\)

Step 2: Factor the quadratic expression \(x^2 + 2x + 1\) as a perfect square:

\(x^2 + 2x + 1 = (x + 1)^2\)

Step 3: Final factorization:

\(x^3 - 3x^2 - 9x - 5 = (x - 5)(x + 1)^2\)

Both methods yield the same factorization: \(x^3 - 3x^2 - 9x - 5 = (x^2 - 5)(x - 1)^2\).

Verified answer

Answer:

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

Explanation:

Given expression is

[tex]\bf \: {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]\bf\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{\sf{\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]