Problem 48 - Precalculus Mathematics for Calculus

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\). $$P(x)=x^{3}-5 x^{2}+2 x+12$$

Step-by-Step Solution

Answer

The real zeros are \(x = 3\), \(x = 1 + \sqrt{5}\), and \(x = 1 - \sqrt{5}\).

Step 1: Identify Possible Rational Zeros

According to the Rational Root Theorem, any rational zero of \(P(x)\) must be a factor of the constant term (12) divided by a factor of the leading coefficient (1). Thus, the possible rational zeros are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\).

Step 2: Test Possible Rational Zeros

We will use synthetic division or direct substitution to test these possible zeros. First, let's test \(x = 1\):\[P(1) = 1^3 - 5 \cdot 1^2 + 2 \cdot 1 + 12 = 1 - 5 + 2 + 12 = 10\] Since \(P(1) eq 0\), \(x = 1\) is not a zero. Let's try \(x = 2\):\[P(2) = 2^3 - 5 \cdot 2^2 + 2 \cdot 2 + 12 = 8 - 20 + 4 + 12 = 4\]Since \(P(2) eq 0\), \(x = 2\) is not a zero. Now test \(x = 3\):\[P(3) = 3^3 - 5 \cdot 3^2 + 2 \cdot 3 + 12 = 27 - 45 + 6 + 12 = 0\]\(x = 3\) is a zero!

Step 3: Perform Polynomial Division

Divide \(P(x)\) by \(x - 3\) using synthetic division to factor the polynomial.Carry out the synthetic division with \(x = 3\):\(| \begin{array}{r|rrr}3 & 1 & -5 & 2 & 12 \ & & 3 & -6 & -12 \\hline & 1 & -2 & -4 & 0 \\end{array} |\)This results in the polynomial \(x^2 - 2x - 4\). Thus, \(P(x) = (x - 3)(x^2 - 2x - 4)\).

Step 4: Solve the Quadratic Equation

The factor \(x^2 - 2x - 4 = 0\) is a quadratic equation. Solve it using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here \(a = 1\), \(b = -2\), \(c = -4\):\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\]\[x = \frac{2 \pm \sqrt{4 + 16}}{2}\]\[x = \frac{2 \pm \sqrt{20}}{2}\]\[x = \frac{2 \pm 2\sqrt{5}}{2}\]\[x = 1 \pm \sqrt{5}\]

Step 5: List All Real Zeros

The real zeros of the polynomial \(P(x)=x^{3}-5x^{2}+2x+12\) are \(x = 3\), \(x = 1 + \sqrt{5}\), and \(x = 1 - \sqrt{5}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is an essential tool for finding rational zeros of a polynomial. It helps narrow down the list of potential rational solutions. Here's how it works: For any polynomial with integer coefficients, the possible rational roots are all the factors of the constant term divided by the factors of the leading coefficient.
In the case of the polynomial \(P(x) = x^3 - 5x^2 + 2x + 12\), the constant term is 12, and the leading coefficient is 1. Therefore, we need to find all factors of 12, which are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). These give us the potential rational zeros to test.
This theorem is particularly useful because it significantly reduces the number of possible rational numbers we need to test to see if they are indeed zeros of the polynomial.
Once you identify potential rational roots, you can use other methods, like synthetic division, to test them.

Synthetic Division

Synthetic Division is a simplified form of polynomial division, especially handy when testing potential roots. Once you have identified a possible zero using the Rational Root Theorem, synthetic division helps verify it.
When we use synthetic division for the polynomial \(P(x) = x^3 - 5x^2 + 2x + 12\) and test \(x = 3\), we're essentially trying to see if dividing \(P(x)\) by \(x-3\) yields a remainder of zero. If it does, then \(x=3\) is confirmed as a root.
Here's a brief outline:

• Write down the coefficients of \(P(x)\): 1, -5, 2, and 12.
• Bring down the first coefficient.
• Multiply by the test root (3) and add to the next coefficient.
• Repeat till the end.

For our specific case after synthetic division, the polynomial becomes \(x^2 - 2x - 4\), with a remainder of 0, confirming \(x = 3\) is a root of the polynomial.

Quadratic Formula

The Quadratic Formula is used to find the roots of a quadratic equation. After simplifying \(P(x)\) using synthetic division, the resulting quadratic factor is \(x^2 - 2x - 4 = 0\).
For a quadratic equation \(ax^2 + bx + c = 0\), the roots can be found using:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plugging in the values from the quadratic part of our polynomial:

• \(a = 1\)
• \(b = -2\)
• \(c = -4\)

The formula becomes:\[x = \frac{2 \pm \sqrt{20}}{2}\]Further simplifying gives the zeros \(x = 1 \pm \sqrt{5}\).
These represent the points where the quadratic part of \(P(x)\) crosses the x-axis, alongside our previously identified root, \(x=3\). These zeros are crucial for understanding the behavior and graph of any polynomial.