Problem 39 - College Algebra with Modeling and Visualization

Question

Solve the polynomial equation. $$ x^{4}+x^{3}=16-8 x-6 x^{2} $$

Step-by-Step Solution

Answer

The solutions are \(x = -2\), \(x = 2\), and \(x = \frac{-1 \pm i\sqrt{15}}{2}\).

Step 1: Rearrange the Equation

Start by moving all the terms to one side to set the equation to zero: Move the terms on the right side to the left: \(x^4 + x^3 + 6x^2 + 8x - 16 = 0\)

Step 2: Identify Potential Rational Roots

Use the Rational Root Theorem to identify potential rational roots. The potential roots are the factors of the constant term (\(-16\)) divided by the factors of the leading coefficient (\(1\)).The factors of \(-16\) are: \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\).So, the possible rational roots are: \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\).

Step 3: Test Potential Roots Using Synthetic Division

Test these potential roots using synthetic division to identify actual roots.First, test \(x = 2\): Using synthetic division:\[\begin{array}{r|rrrr}2 & 1 & 1 & 6 & 8 & -16 \ & & 2 & 6 & 24 & 64 \\hline & 1 & 3 & 12 & 32 & 48\end{array}\]The remainder is not zero. Therefore, \(x = 2\) is not a root.Now, test \(x = -2\):\[\begin{array}{r|rrrr}-2 & 1 & 1 & 6 & 8 & -16 \ & & -2 & 2 & -16 & 16 \\hline & 1 & -1 & 4 & -8 & 0\end{array}\]The remainder is zero. Therefore, \(x = -2\) is a root.

Step 4: Factor the Polynomial

Since \(x = -2\) is a root, \((x + 2)\) is a factor. The quotient from synthetic division is \(x^3 - x^2 + 4x - 8\).Therefore, the equation can be rewritten as:\((x + 2)(x^3 - x^2 + 4x - 8) = 0\).

Step 5: Solve the Cubic Polynomial

To solve the cubic polynomial \(x^3 - x^2 + 4x - 8 = 0\), start by checking for rational roots again.We test \(x = 2\) with the same method:\[\begin{array}{r|rrr}2 & 1 & -1 & 4 & -8 \ & & 2 & 2 & 12 \\hline & 1 & 1 & 6 & -8\end{array}\]The remainder is zero, so \(x = 2\) is a root for the cubic polynomial. Thus \((x - 2)\) is a factor.The quotient is \(x^2 + x + 4\).

Step 6: Solve the Quadratic Equation

Now, we solve \(x^2 + x + 4 = 0\) using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = 1\), \(c = 4\).\[x = \frac{-1 \pm \sqrt{1 - 16}}{2}\]\[x = \frac{-1 \pm \sqrt{-15}}{2}\]The solutions are complex: \(x = \frac{-1 + i\sqrt{15}}{2}\) and \(x = \frac{-1 - i\sqrt{15}}{2}\).

Step 7: Compile the Solutions

From the previous steps, the roots of the polynomial equation are:1. \(x = -2\)2. \(x = 2\)3. \(x = \frac{-1 + i\sqrt{15}}{2}\)4. \(x = \frac{-1 - i\sqrt{15}}{2}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for solving polynomial equations. It helps you find possible rational solutions to the equation. Basically, it suggests that any rational root, expressed as a fraction \(\frac{p}{q}\), must have a numerator \(p\) that is a factor of the constant term and a denominator \(q\) that is a factor of the leading coefficient.
To put it simply, if you have a polynomial \(ax^n + bx^{n-1} + ... + k = 0\), based on the Rational Root Theorem, you generate the potential rational roots by dividing the factors of \(k\) (constant term) by the factors of \(a\) (leading coefficient).
In the exercise you worked on, the equation \(x^4 + x^3 + 6x^2 + 8x - 16 = 0\) had the leading coefficient of 1 and a constant of \(-16\). This made the possible rational roots easy to identify: \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\). These are tested further to find the actual roots.

Synthetic Division

Synthetic division is a simpler and faster way to divide polynomials, compared to the long division method. It's particularly useful when dividing a polynomial by a binomial of the form \((x - c)\).
Here's a brief overview of how it's done:

• Write down the coefficients of the polynomial.
• Place the value of \(c\) (from \(x - c\)) to the left.
• Bring down the first coefficient to the bottom row.
• Multiply this number by \(c\) and add it to the next coefficient.
• Repeat this process until all coefficients are used.

If the final number (remainder) is zero, then \(c\) is indeed a root of the polynomial.
When using this method for your polynomial, you tested roots like \(x = 2\) and \(x = -2\) to locate which values would result in a remainder of zero, confirming them as roots. This step simplifies calculating and reveals factors like \((x + 2)\) and \((x - 2)\).

Quadratic Formula

When faced with a quadratic equation that cannot be easily factored, the quadratic formula offers a foolproof solution method. The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This equation allows you to find the roots of any quadratic equation \(ax^2 + bx + c = 0\).
Here's what the components mean:

• \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation.
• The expression under the square root, \(b^2 - 4ac\), is called the discriminant.

The discriminant helps you figure out the nature of the roots:

• If it's positive, the equation has two distinct real roots.
• If it's zero, there's exactly one real root.
• If it's negative, the roots are complex (involving imaginary numbers).

In your exercise, the quadratic \(x^2 + x + 4 = 0\) was solved using this formula, resulting in the complex roots \(x = \frac{-1 + i\sqrt{15}}{2}\) and \(x = \frac{-1 - i\sqrt{15}}{2}\). These outcomes show the power of the quadratic formula in providing precise answers.