Problem 64 - Precalculus Mathematics for Calculus

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\) $$P(x)=x^{5}-x^{4}-6 x^{3}+14 x^{2}-11 x+3$$

Step-by-Step Solution

Answer

The real zeros are \(x = 1\) (double root), \(x = 3\), \(x = 1 + \sqrt{2}\), \(x = 1 - \sqrt{2}\).

Step 1: Identify Potential Rational Roots

Using the Rational Root Theorem, we identify potential rational roots of the polynomial \(P(x) = x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3\). The possible rational roots are \(\pm 1, \pm 3\), as these are the factors of the constant term (3) divided by the factors of the leading coefficient (1).

Step 2: Test Possible Rational Roots

We test the potential roots using synthetic division or direct substitution into \(P(x)\). After testing \(x = 1\) and \(x = 3\), we find that \(x = 1\) is a root, because \(P(1) = 0\). So, \(x-1\) is a factor of the polynomial.

Step 3: Perform Polynomial Division

Divide \(P(x)\) by \(x - 1\) using synthetic division. This yields the quotient \(x^4 - 6x^3 + 8x^2 + 6x - 3\). Thus, we have: \(P(x) = (x - 1)(x^4 - 6x^3 + 8x^2 + 6x - 3)\).

Step 4: Factor the Quartic Polynomial

To find the zeros of the quartic polynomial \(x^4 - 6x^3 + 8x^2 + 6x - 3\), we apply the Rational Root Theorem again and find that \(x = 3\) is a root. We then perform synthetic division on \(x^4 - 6x^3 + 8x^2 + 6x - 3\) by \(x-3\), giving us \((x-3)(x^3 - 3x^2 - x + 1)\).

Step 5: Factor the Cubic Polynomial Further

Analyze \(x^3 - 3x^2 - x + 1\) for rational roots. After testing possible rational roots, we find that \(x = 1\) works again. Perform synthetic division to reduce \(x^3 - 3x^2 - x + 1\) by \(x-1\), which gives \((x - 1)(x^2 - 2x - 1)\). The expression now becomes \((x - 1)^2(x - 3)(x^2 - 2x - 1)\).

Step 6: Solve the Quadratic Equation

The quadratic \(x^2 - 2x - 1 = 0\) does not factor easily, so we use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a=1\), \(b=-2\), and \(c=-1\), the solutions are \(x = 1 \pm \sqrt{2}\).

Step 7: List All Real Zeros

Combine all discovered zeros. The real zeros are \(x = 1\) (with multiplicity 2), \(x = 3\), \(x = 1 + \sqrt{2}\), and \(x = 1 - \sqrt{2}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for solving polynomial equations. It helps us guess potential rational zeros of a polynomial by examining its structure.

Here's how it works:

• The polynomial's constant term and the leading coefficient's factors are taken.
• We then form all possible fractions using factors of the constant term as numerators and factors of the leading coefficient as denominators.
• These fractions are the potential rational roots of the polynomial.

In the exercise, we derived the possible rational roots for the polynomial \(P(x) = x^5 - x^4 - 6x^3 + 14x^2 - 11x + 3\) by considering the factors of 3 (constant term) and 1 (leading coefficient). Thus, the possible rational roots were \(\pm 1, \pm 3\).

Using the Rational Root Theorem gives us valuable starting points to check for actual zeros.

Synthetic Division

Once we suspect a potential zero, we can use synthetic division to verify it. Synthetic division offers a streamlined alternative to polynomial long division. It tells us not only if our suspected zero works, but also helps in factoring the polynomial.

Here's how it progresses:

• Write down the coefficients of the polynomial.
• Pick a suspected root and use it in the division process.
• Perform operations similar to long division but quicker.
• The remainder reveals whether the suspected root is indeed a zero.

In our scenario, we used synthetic division first to test \(x = 1\) and later \(x = 3\), discovering that \(x=1\) is a root. This helped us factor our initial quintic polynomial down to a quartic, and eventually further down, simplifying it into smaller manageable pieces.

Quadratic Formula

When a polynomial reduces to a tough quadratic, the Quadratic Formula is our savior. It gives exact roots for quadratic equations, even those hard to factor.

The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our reduced polynomial, \(x^2 - 2x - 1 = 0\), we saw that factoring wasn’t straightforward. By putting \(a=1\), \(b=-2\), and \(c=-1\) into the formula, the solutions turned out as \(x = 1 \pm \sqrt{2}\).

This formula is a powerful resource, ensuring we confidently find roots where simpler methods might stumble.

Multiplicity of Roots

Roots can appear more than once in a polynomial and this is known as their multiplicity.

Recognizing multiplicity is crucial because:

• It influences the nature of the polynomial's graph - whether it just touches or crosses the x-axis at that root.
• It impacts the total count of the polynomial's degree.

In the exercise at hand, \(x = 1\) appeared twice which told us it has a multiplicity of 2. It's reflected in the factor \((x-1)^2\) being part of the polynomial equation.

Understanding the multiplicity helps us comprehend the complete solution and behavior of polynomials.