Factorize: x³ - 3x² - 9x - 5

To factor the polynomial \(x^3 - 3x^2 - 9x - 5\), we can start by looking for rational roots using the Rational Root Theorem. This theorem suggests that any potential rational root, in the form of \(p/q\), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient, could be a candidate for testing. In this case, the constant term is \(-5\) and the leading coefficient is \(1\).

Identifying Potential Roots

The factors of \(-5\) are \(\pm 1\), \(\pm 5\). Since the leading coefficient is \(1\), the possible rational roots to test are \(-5\), \(-1\), \(1\), and \(5\).

Testing the Roots

Let's evaluate these candidates by substituting them into the polynomial:

• For \(x = 1\):

  \(1^3 - 3(1^2) - 9(1) - 5 = 1 - 3 - 9 - 5 = -16\) (not a root)

• For \(x = -1\):

  \((-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0\) (this is a root)

• For \(x = 5\):

  \(5^3 - 3(5^2) - 9(5) - 5 = 125 - 75 - 45 - 5 = 0\) (this is also a root)

• For \(x = -5\):

  \((-5)^3 - 3(-5)^2 - 9(-5) - 5 = -125 - 75 + 45 - 5 = -160\) (not a root)

Factoring the Polynomial

Since we found that \(x = -1\) is a root, we can use synthetic division to divide the polynomial by \(x + 1\).

Synthetic Division

Setting up synthetic division with the coefficients \(1, -3, -9, -5\):

• Write down the coefficients: \(1, -3, -9, -5\)
• Use \(-1\) (the root) for synthetic division:

-1 |  1   -3   -9   -5
   |      -1    4    5
-----------------------
     1   -4   -5   0

The result of the division is \(x^2 - 4x - 5\). Thus, we can express the original polynomial as:

\(x^3 - 3x^2 - 9x - 5 = (x + 1)(x^2 - 4x - 5)\)

Further Factoring

Next, we can factor \(x^2 - 4x - 5\). We look for two numbers that multiply to \(-5\) and add to \(-4\). The numbers \(-5\) and \(1\) fit this requirement:

So, we can factor \(x^2 - 4x - 5\) as \((x - 5)(x + 1)\).

Final Factorization

Putting it all together, we have:

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

This can be simplified to:

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

Summary

The complete factorization of the polynomial \(x^3 - 3x^2 - 9x - 5\) is \((x + 1)^2 (x - 5)\). This means that the polynomial has a double root at \(x = -1\) and a single root at \(x = 5\).