Problem 77 - A Graphical Approach to Precalculus with Limits

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=x^{3}+5 x^{2}-3 x-15 ; \quad k=-5$$

Step-by-Step Solution

Answer

The polynomial factors as \( (x + 5)(x - \sqrt{3})(x + \sqrt{3}) \).

Step 1: Verify Given Zero

First, we'll confirm that the given zero \( k = -5 \) is indeed a zero of \( P(x) \). Substitute \( x = -5 \) into \( P(x) \) and check if the result is zero. \[ P(-5) = (-5)^3 + 5(-5)^2 - 3(-5) - 15 = -125 + 125 + 15 - 15 = 0 \] Since \( P(-5) = 0 \), \( k = -5 \) is a zero of \( P(x) \).

Step 2: Divide by (x + 5)

Since \( -5 \) is a zero, \( x + 5 \) is a factor of \( P(x) \). Use polynomial division to divide \( P(x) \) by \( x + 5 \) to obtain the other factors.\[ \begin{array}{r|rr} & x^2 & 0x & -3 \hline x + 5 & x^3 & +5x^2 & -3x & -15 \ & -(x^3 + 5x^2) \ & 0 & -3x & -15 \ & +3(x + 5)\ & -(-3x - 15) \ & 0 \end{array} \]The quotient is \( x^2 - 3 \).

Step 3: Factor the Quotient

Factor the quadratic \( x^2 - 3 \). This is a difference of squares and can be written as \( (x - \sqrt{3})(x + \sqrt{3}) \).

Step 4: Write the Factored Form

The complete factorization of \( P(x) \) into linear factors is given by: \[ P(x) = (x + 5)(x - \sqrt{3})(x + \sqrt{3}) \] This represents \( P(x) \) completely factored into linear components.

Key Concepts

Linear Factors

When we talk about factoring a polynomial into linear factors, we're aiming to express the polynomial as a product of factors of the form \( x - a \), where \( a \) is a root or zero of the polynomial. Each linear factor corresponds to roots or solutions of the polynomial equation.
For instance, in the polynomial \( P(x) = x^3 + 5x^2 - 3x - 15 \), after finding that \( k = -5 \) is a zero, \( x + 5 \) is identified as one of the linear factors. This is because when \( x = -5 \), the polynomial equals zero.
The process of breaking down \( P(x) \) into these simpler components lets us understand the behavior and solutions of the polynomial more clearly. Ultimately, finding linear factors is like opening a lock: each factor is a key to the solutions or roots of the polynomial.

Zeros of Polynomials

The zeros of a polynomial are values of \( x \) that make the polynomial equal to zero. They are also referred to as roots or solutions. Finding these zeros is a fundamental aspect of solving polynomial equations.
In the exercise, \( -5 \) is given as a zero of \( P(x) = x^3 + 5x^2 - 3x - 15 \). By substituting \( x = -5 \) into the polynomial and obtaining zero, we confirm that \( -5 \) is indeed a zero. This confirmation process is crucial because it forms the basis for further factorizations or simplifications.
Zeros reveal where a polynomial graph will intersect the x-axis, and each zero corresponds to at least one linear factor. For our example, knowing \( -5 \) is a zero allowed us to initially factor \( P(x) \) by \( x + 5 \). Identifying all zeros fully factors the polynomial and provides all solutions to the polynomial equation.

Polynomial Division

Polynomial division comes into play when we need to divide a polynomial to simplify it or to find its factors. It's similar to long division performed with numbers but involves algebraic expressions.
In our example, once we knew \( -5 \) was a zero, we used polynomial division to divide \( P(x) = x^3 + 5x^2 - 3x - 15 \) by \( x + 5 \). What remains after the division is called the quotient. This quotient helps in further factoring, as seen with the result \( x^2 - 3 \).
Polynomial division may seem intricate at first, but it's a powerful tool. By simplifying complex polynomials into simpler forms, it helps identify all possible linear factors and zeros. This step is essential for a comprehensive understanding of a polynomial's structure and solutions.