Question

if \\(f(1) = 0\\), what are all the roots of the function \\(f(x) = x^3 + 3x^2 - x - 3\\)? use the remainder theorem.

\\(x = -1, x = 1, \text{ or } x = 3\\)
\\(x = -3, x = -1, \text{ or } x = 1\\)
\\(x = -3 \text{ or } x = 1\\)
\\(x = -1 \text{ or } x = 3\\)

Explanation:

🆕 New Concept Discovered: Polynomial Factor and Remainder Theorems
Using remainders to find factors and solve higher-degree equations.

Step 1: Identify the first root and factor

Since we are given that \( f(1) = 0 \), the Remainder Theorem (and Factor Theorem) tells us that \( x = 1 \) is a root of the function, and \( (x - 1) \) is a factor of:

$$ f(x) = x^3 + 3x^2 - x - 3 $$

Step 2: Divide the polynomial to find the remaining quadratic factor

We divide \( f(x) \) by \( (x - 1) \) using synthetic division:

$$
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$$

The quotient is the quadratic expression:

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

Step 3: Factor the quadratic expression to find the remaining roots

Set the quadratic quotient to zero to find the other roots:

$$ x^2 + 4x + 3 = 0 $$

Factor the trinomial:

$$ (x + 3)(x + 1) = 0 $$

This gives the remaining roots:

$$ x = -3 \quad \text{and} \quad x = -1 $$

Combining these with our first root \( x = 1 \), the complete set of roots is:

$$ x = -3, \quad x = -1, \quad \text{or} \quad x = 1 $$

Answer:

\( x = -3, x = -1, \text{ or } x = 1 \)