Question

f(x) = x^3 - 2x^2 - 7x - 4

Explanation:

Identify the given function

We are given the cubic polynomial function:

$$f(x) = x^3 - 2x^2 - 7x - 4$$

Our goal is to find its roots (or zeros) by factoring the polynomial completely.

Find a rational root

We test potential integer roots using the Rational Root Theorem. The factors of the constant term \(-4\) are \(\pm 1, \pm 2, \pm 4\).
Let's test \(x = -1\):

$$f(-1) = (-1)^3 - 2(-1)^2 - 7(-1) - 4$$

$$f(-1) = -1 - 2 + 7 - 4 = 0$$

Since \(f(-1) = 0\), \(x = -1\) is a root, which means \((x + 1)\) is a factor.

Perform polynomial division

We divide \(f(x)\) by \((x + 1)\) using synthetic division or polynomial long division to find the remaining quadratic factor:

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The quotient is the quadratic expression:

$$q(x) = x^2 - 3x - 4$$

Factor the quadratic quotient

Now we factor the quadratic expression \(x^2 - 3x - 4\). We look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\):

$$x^2 - 3x - 4 = (x - 4)(x + 1)$$

Write the fully factored form and find all roots

Combining all the factors, we get:

$$f(x) = (x + 1)(x + 1)(x - 4) = (x + 1)^2(x - 4)$$

Setting \(f(x) = 0\) gives the roots:

$$x = -1 \quad (\text{multiplicity } 2)$$

$$x = 4 \quad (\text{multiplicity } 1)$$

Answer:

The fully factored form of the polynomial is:

$$f(x) = (x + 1)^2(x - 4)$$

The roots of the function are:

$$x = -1 \quad \text{and} \quad x = 4$$