Problem 37 - Precalculus Mathematics for Calculus

Question

Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$P(x)=x^{4}-2 x^{3}-8 x+16$$

Step-by-Step Solution

Answer

Zeros are \(x = 2\) (double root) and complex roots \(-1 \pm i\sqrt{3}\). The graph is tangent to the x-axis at \(x = 2\).

Step 1: Check for Common Factors

First, check if there is a common factor for all terms in the polynomial \(P(x) = x^{4} - 2x^{3} - 8x + 16\). There isn't a common factor for all terms.

Step 2: Apply Rational Root Theorem

Use the Rational Root Theorem to determine possible rational roots. The potential rational roots are the factors of the constant term \(16\), which are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\). Evaluate these in the polynomial to test which are actual roots.

Step 3: Evaluate Polynomial to Find a Root

Test the potential roots from Step 2. We find that \(P(2) = 0\). This means \(x = 2\) is a root, thus \((x - 2)\) is a factor.

Step 4: Perform Synthetic Division

Use synthetic division to divide the polynomial \(P(x)\) by \((x-2)\) to find the quotient polynomial. This results in \(x^{3} - 8\).

Step 5: Factor Quotient Further if Possible

The quotient polynomial \(x^3 - 8\) is recognized as a difference of cubes. It can be factored using the formula \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\) where \(a = x\) and \(b=2\). This gives \(x^3 - 8 = (x-2)(x^2 + 2x + 4)\).

Step 6: Write Complete Factored Form

Combine the factors from Steps 3 and 5 to get the complete factored form of the original polynomial: \(P(x) = (x-2)^2(x^2 + 2x + 4)\).

Step 7: Find Zeros from Factored Form

To find the zeros, set each factor equal to zero. For \((x-2)^2 = 0\), we get \(x = 2\). For \(x^2 + 2x + 4 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=2, c=4\) to find the solutions.

Step 8: Calculate Quadratic Zeros

Calculate the roots of \(x^2 + 2x + 4 = 0\) using the quadratic formula: \(x = \frac{-2 \pm \sqrt{4 - 16}}{2}\). The discriminant is negative \(-12\), indicating complex roots. Thus, the roots are \(-1 + i\sqrt{3}\) and \(-1 - i\sqrt{3}\).

Step 9: Sketch the Graph with Zeros

Plot the zeros on the graph. There is a double root at \(x = 2\) with a multiplicity of 2, which makes the graph tangent to the \(x\)-axis at \(x=2\). There are no real zeros from \(x^2 + 2x + 4\), so no additional x-intercepts.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a beneficial tool when it comes to factoring polynomials and finding their roots. It suggests that if a polynomial has a rational root \(\frac{p}{q}\), then \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. For the polynomial \(P(x) = x^4 - 2x^3 - 8x + 16\), identify the constant term, which is 16, and the leading coefficient, which is 1.

• The factors of 16 are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\).
• The leading coefficient is 1, so its only factor is \(\pm 1\).
• This means the possible rational roots are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\).

Next, evaluate these potential roots in the polynomial to check which ones yield a zero. This step can save time as it helps narrow down the actual roots, making your work with polynomials more efficient.

Synthetic Division

Once a potential root is found using the Rational Root Theorem, synthetic division offers a fast way to divide the polynomial by \((x-r)\) to obtain the quotient. Let's say we discover that \(x=2\) is a root for \(P(x) = x^4 - 2x^3 - 8x + 16\).

• Write the coefficients of \(P(x)\) in a row: 1, -2, 0, -8, 16.
• Perform synthetic division using the found root, \(x=2\).
• The process involves repeated multiplication and addition.
• In this case, it results in a quotient of \(x^3 - 8\).

This method is quicker and easier than long division, especially for polynomials with higher degrees. It simplifies the process of finding all factors of the polynomial.

Complex Roots

When factoring polynomials, some equations might lead to complex roots. After applying synthetic division to the polynomial \(P(x) = x^4 - 2x^3 - 8x + 16\), we find the quotient \(x^3 - 8\), which can be further factored into \((x-2)(x^2 + 2x + 4)\).
To find the roots of \(x^2 + 2x + 4\), employ the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• Here, \(a=1\), \(b=2\), and \(c=4\).
• The discriminant \(b^2 - 4ac = 4 - 16 = -12\) is negative, indicating complex roots.
• Solving, the roots are \(-1 + i\sqrt{3}\) and \(-1 - i\sqrt{3}\).

Complex roots appear in conjugate pairs and suggest that the graph of the polynomial will not intersect the x-axis at these values. Understanding how and why complex roots form is crucial as they reveal that a polynomial cannot always be factored into real-number solutions.