Problem 37 - Basic Engineering Mathematics

Question

Factorize \(x^{3}+3 x^{2}-x-3\)

Step-by-Step Solution

Answer

The factorization is \((x - 1)(x + 1)(x + 3)\).

Step 1: Identify Potential Roots Using the Rational Root Theorem

The Rational Root Theorem suggests that any rational root of the polynomial \(x^3 + 3x^2 - x - 3\) is a divisor of the constant term, -3. The divisors of -3 are \(\pm 1\) and \(\pm 3\). These will be our potential roots.

Step 2: Test Potential Roots

Substitute the potential roots into the polynomial to determine which, if any, are actual roots.1. For \(x = 1\), we have: \[f(1) = 1^3 + 3(1)^2 - 1 - 3 = 0\] \(x = 1\) is a root, so \((x - 1)\) is a factor.2. None of the other potential roots, \(x = -1\), \(x = 3\), or \(x = -3\), yield \(f(x) = 0\). So, they are not roots.

Step 3: Divide the Polynomial by \(x - 1\)

Since \(x = 1\) is a root, divide \(x^3 + 3x^2 - x - 3\) by \(x - 1\) using synthetic division or long division.1. Dividing, we get: \[ \begin{array}{c|cccc} 1 & 1 & 3 & -1 & -3 \ & & 1 & 4 & 3 \ \hline & 1 & 4 & 3 & 0 \ \end{array} \] The quotient is \(x^2 + 4x + 3\) with no remainder, so \(x^3 + 3x^2 - x - 3 = (x - 1)(x^2 + 4x + 3)\).

Step 4: Factor the Quadratic Polynomial

The next step is to factor the quadratic \(x^2 + 4x + 3\).The factors of 3 that add up to 4 are 1 and 3. Thus:\[x^2 + 4x + 3 = (x + 1)(x + 3)\].

Step 5: Write the Complete Factorization

Combine the results from previous steps. The complete factorization of \(x^3 + 3x^2 - x - 3\) using the found factors is:\[(x - 1)(x + 1)(x + 3)\].

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool when working with polynomial equations, especially when you want to find rational solutions quickly. Given a polynomial with integer coefficients, this theorem suggests that any rational root, in its simplest form, is a fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient.
In our exercise, we have the polynomial \( x^3 + 3x^2 - x - 3 \), where the constant term is \(-3\) and the leading coefficient is \(1\). This means any rational root of this polynomial must be a divisor of \(-3\).

• List of divisors for \(-3\): \( \pm 1 \), \( \pm 3 \)
• These values serve as potential candidates for rational roots.

By systematically checking these potential roots through substitution, we ascertain which, if any, are actual roots of the polynomial. This process is efficient and often the first step in factorization.

Synthetic Division

Synthetic Division is an efficient method for dividing a polynomial by a linear factor of the form \( x - a \). It's typically quicker than long division and is especially useful when testing potential roots gained from the Rational Root Theorem.
Here's how it works with our example: Once we've identified \( x = 1 \) as a root, we use synthetic division to divide \( x^3 + 3x^2 - x - 3 \) by \( x - 1 \).

• Write down the coefficients of the polynomial: 1, 3, -1, -3.
• Place the root (1 in this case) outside the division symbol.
• Bring down the first coefficient as it is.
• Multiply the root by this number, place it under the next coefficient, and add these two numbers.

This process is repeated across the row, simplifying calculations and leading us to the quotient polynomial \( x^2 + 4x + 3 \), confirming that \( x - 1 \) is a factor because there's no remainder.

Quadratic Polynomial

A quadratic polynomial is a second-degree polynomial that can often be factored into two binomials. After using synthetic division, we're left with the quadratic \( x^2 + 4x + 3 \). The goal here is to factor this further.
To factor a quadratic of the form \( ax^2 + bx + c \), look for two numbers that multiply to \( c \) and add to \( b \). For our quadratic:

• We need two numbers that multiply to 3 (constant term) and add up to 4 (coefficient of \( x \)).
• The numbers 1 and 3 multiply to 3 and add up to 4, making them the correct choices.

Thus, the quadratic \( x^2 + 4x + 3 \) factors to \( (x + 1)(x + 3) \).
This breakdown completes the factorization of the original polynomial, as we now have \((x - 1)(x + 1)(x + 3)\) as the fully factored form of \( x^3 + 3x^2 - x - 3 \). This understanding reinforces the factorization process and confirms accuracy, which is crucial in polynomial factorization.