Problem 16 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ 2 x^{3}+x^{2}-18 x-9=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 3\), \(x = -1\), and \(x = -\frac{3}{2}\).

Step 1: Identify Possible Rational Roots

Use the Rational Root Theorem to identify possible rational roots. According to this theorem, possible rational roots are factors of the constant term (-9) divided by factors of the leading coefficient (2). Hence, the possible rational roots are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2} \).

Step 2: Test Possible Roots

Substitute possible rational roots into the equation to see if they satisfy the equation \(2x^3 + x^2 - 18x - 9 = 0\). After testing, \(x = 3\) satisfies the equation as it results in zero.

Step 3: Perform Synthetic Division

Use synthetic division to divide the cubic polynomial \(2x^3 + x^2 - 18x - 9\) by \(x - 3\). This results in a quotient polynomial \(2x^2 + 7x + 3\).

Step 4: Solve the Quadratic Equation

The quotient from the previous step is \(2x^2 + 7x + 3 = 0\). Solve this quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=2\), \(b=7\), and \(c=3\).

Step 5: Calculate Quadratic Formula Results

Substitute into the quadratic formula: \(x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot 3}}{4}\) which simplifies to \(x = \frac{-7 \pm \sqrt{25}}{4}\). This gives the roots \(x = -1\) and \(x = -\frac{3}{2}\).

Step 6: List All Real Solutions

The real roots found are \(x = 3\), \(x = -1\), and \(x = -\frac{3}{2}\). So, the original polynomial equation has three real solutions.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a fantastic tool for determining possible rational solutions to a polynomial equation. When you're faced with a polynomial equation, like our example: \[2x^3 + x^2 - 18x - 9 = 0\],this theorem allows us to guess our first steps toward the solution.

• First, look at the constant term (the number without a variable) and the leading coefficient (the coefficient of the highest power of the variable).
• In our example, the constant term is \(-9\), and the leading coefficient is \(2\).
• According to the Rational Root Theorem, any rational root of this polynomial, when expressed in its simplest form, is of the form\( \pm \frac{p}{q} \), where \(p\) represents factors of the constant term and \(q\) represents factors of the leading coefficient.
• For our equation, factors of \(-9\) are \(\pm 1, \pm 3,\), and \(\pm 9\), and factors of \(2\) are \(\pm 1\), and \(\pm 2\).

This means the potential rational roots are \(\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}\). These roots are starting points for testing if any of them satisfy our polynomial.

Synthetic Division

After identifying possible rational roots, synthetic division is used to verify them. It provides a simplified way to divide polynomials and helps to test these roots quickly.
Let's break down how synthetic division works, using our equation \(2x^3 + x^2 - 18x - 9 = 0\) and the root \(x = 3\):

• Write down the coefficients of the polynomial in order. For our equation, it would be \(2, 1, -18, -9\).
• Place the root you are testing (here, \(3\)) outside the synthetic division bracket.
• Bring the leading coefficient (\(2\)) directly down.
• Multiply this result by the root \(3\) and write it under the next coefficient (\(1\)).
• Add this product to the next coefficient and write the result below.
• Continue this process across all coefficients.

If the remainder equals zero, as it does in this case, \(3\) is indeed a root, and you get a quotient polynomial. Here, it simplifies the original polynomial to \(2x^2 + 7x + 3\). This new polynomial can then be solved to find other roots.

Quadratic Formula

When dealing with polynomials, reducing them potentially gets you to a quadratic equation, which can be solved using the Quadratic Formula. This formula is a go-to method when other factoring isn't straightforward.
The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]and it applies to any quadratic equation in the form \(ax^2 + bx + c = 0\).
For our remaining polynomial from synthetic division, \(2x^2 + 7x + 3\), the coefficients are \(a = 2, b = 7, c = 3\).

• Insert these values into the quadratic formula:\[x = \frac{-7 \pm \sqrt{7^2 - 4 \times 2 \times 3}}{4}\]
• Calculate the discriminant \(\sqrt{49 - 24}\) first, resulting in \(\sqrt{25}\), which simplifies to \(5\).
• Using these, solve for \(x\), giving two solutions as:\[x = \frac{-7 + 5}{4} = -1\]\[x = \frac{-7 - 5}{4} = -\frac{3}{2}\].

This process reveals two additional real roots for the polynomial, adding to the root we found earlier using synthetic division. So, the complete set of real solutions for the original equation is \(x = 3, x = -1, x = -\frac{3}{2}\).