Problem 56 - A Graphical Approach to Precalculus with Limits

Question

The close relationships among \(x\) -intercepts of a graph of a function, real zeros of the function, and real solutions of the associated equation should, by now. be apparent to you. Consider the graph of the polynomial function \(P\), given by \(Y_{1}=X^{3}-2 X^{2}-11 X+12.\) (GRAPH CANT COPY) What are the zeros of the function \(P ?\)

Step-by-Step Solution

Answer

The zeros are \( x = 1 \), \( x = 4 \), and \( x = -3 \).

Step 1: Understanding the Problem

We need to find the zeros of the polynomial function \( P(x) = x^3 - 2x^2 - 11x + 12 \), which correspond to the \( x \)-intercepts of its graph. These zeros are the values of \( x \) for which \( P(x) = 0 \).

Step 2: Checking Possible Rational Zeros

Use the Rational Root Theorem to identify potential rational zeros. According to the theorem, possible rational zeros are \( \pm \) factors of the constant term (12) divided by the leading coefficient (1). So, the potential rational zeros are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

Step 3: Applying Synthetic Division

Test each potential zero using synthetic division to verify if they are actual zeros of \( P(x) \). Start with \( x = 1 \), \( x = -1 \), \( x = 2 \), etc., and continue until a zero remainder is found.

Step 4: Finding the First Zero with Synthetic Division

After testing, we find\( x = 1 \) is a zero of the function \( P(x) \) using synthetic division:\[\begin{array}{r|rrrr}1 & 1 & -2 & -11 & 12 \ & & 1 & -1 & -12 \\hline & 1 & -1 & -12 & 0 \\end{array}\]The remainder is 0, confirming that \( x = 1 \) is a zero.

Step 5: Factoring the Polynomial

Since \( x = 1 \) is a zero, the polynomial \( P(x) \) can be factored as \( (x - 1)(x^2 - x - 12) \).

Step 6: Solving the Quadratic Factor

Now, solve the quadratic \( x^2 - x - 12 = 0 \) using factoring. This can be factored as \( (x - 4)(x + 3) = 0 \).

Step 7: Finding Additional Zeros

Solve \( (x - 4)(x + 3) = 0 \) using the zero product property:- \( x - 4 = 0 \Rightarrow x = 4 \)- \( x + 3 = 0 \Rightarrow x = -3 \)

Step 8: Conclusion of Zeros

The zeros of the function \( P(x) = x^3 - 2x^2 - 11x + 12 \) are \( x = 1 \), \( x = 4 \), and \( x = -3 \). These are the \( x \)-intercepts.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool for finding the zeros of a polynomial function, especially when dealing with cubic or higher degree polynomials. This theorem states that any rational solution, or root, of a polynomial equation must be a fraction \( \frac{p}{q} \), where:

• \( p \) is a factor of the constant term.
• \( q \) is a factor of the leading coefficient.

In the given polynomial \( P(x) = x^3 - 2x^2 - 11x + 12 \), the constant term is 12 and the leading coefficient is 1. This makes the potential rational roots the divisors of 12: \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
By applying the Rational Root Theorem, you narrow down the possible candidates for potential zeros that you will test using division methods, such as synthetic division.

Synthetic Division

Synthetic division is a convenient method used to determine if a given potential root is indeed a zero of a polynomial. It is simpler and faster than long division, especially useful when testing the numerous possible rational roots that arise from using the Rational Root Theorem.To use synthetic division, follow these steps:

• Write the coefficients of the polynomial in order.
• Choose a potential zero to test (say \( x = 1 \)).
• Carry the first coefficient down.
• Multiply this coefficient by the possible zero, placing the result under the next coefficient.
• Add the numbers in the column, and repeat the above two steps until complete.

Performing synthetic division on our example, testing \( x = 1 \), reveals that it is indeed a zero as the remainder is 0:\[\begin{array}{r|rrrr}1 & 1 & -2 & -11 & 12 \ & & 1 & -1 & -12 \\hline & 1 & -1 & -12 & 0\end{array}\]The zero remainder confirms that \( x = 1 \) is a zero. This result allows us to factor the polynomial further.

Quadratic Formula

The quadratic formula is an essential tool when it comes to solving second-degree polynomials (quadratics) that cannot be easily factored. It provides the roots of any quadratic equation \( ax^2 + bx + c = 0 \) using the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]After factoring out the first zero, \( x = 1 \), from the polynomial \( P(x) \) using synthetic division, the remaining part was a quadratic \( x^2 - x - 12 = 0 \). Fortunately, this quadratic can be factored easily into \( (x - 4)(x + 3) = 0 \).
If factoring wasn't feasible, you'd apply the quadratic formula to find the roots. In our example, no need for the quadratic formula arises, as simple factoring suffices to find \( x = 4 \) and \( x = -3 \).

Polynomial Factoring

Factoring is an effective strategy for solving polynomials by expressing them as a product of simpler polynomials. When you identify one of the roots using tools like the Rational Root Theorem and confirm it with synthetic division, you simplify the original polynomial.For an example like \( P(x) = x^3 - 2x^2 - 11x + 12 \), once a zero (such as \( x = 1 \)) is confirmed through synthetic division, you split the polynomial into two parts: a linear factor \((x-1)\) and the quadratic \(x^2-x-12\).
You then factor the quadratic, resulting in \(x^2 - x - 12 = (x - 4)(x + 3)\). Thus, the entire polynomial is expressed as \((x - 1)(x - 4)(x + 3)\).
Solving each factor separately gives all roots of the polynomial, which are used to identify the polynomial's zeros: \(x = 1, x = 4,\) and \(x = -3\). Factoring turns polynomials into simpler factors, making solving straightforward.