Problem 12 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ 2 x^{3}+x^{2}-18 x-9=0 $$

Step-by-Step Solution

Answer

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

Step 1: Check for Rational Roots

We start by using the Rational Root Theorem to find potential rational solutions. The possible rational roots are the factors of the constant term (-9) divided by the factors of the leading coefficient (2). This gives possible rational roots: \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \). We will test these values by substituting them into the original equation to check if any are roots.

Step 2: Test Rational Root: x = 1

Substitute \( x = 1 \) into the equation: \[2(1)^3 + 1^2 - 18(1) - 9 = 2 + 1 - 18 - 9 = -24 \]This is not zero, so \( x = 1 \) is not a root.

Step 3: Test Rational Root: x = -1

Substitute \( x = -1 \) into the equation:\[2(-1)^3 + (-1)^2 - 18(-1) - 9 = -2 + 1 + 18 - 9 = 8 \]This is not zero, so \( x = -1 \) is not a root.

Step 4: Test Rational Root: x = 3

Substitute \( x = 3 \) into the equation:\[2(3)^3 + 3^2 - 18(3) - 9 = 54 + 9 - 54 - 9 = 0 \]This is zero, so \( x = 3 \) is a root of the equation.

Step 5: Factor the Polynomial using x = 3

Since \( x = 3 \) is a root, we can use polynomial division to factor the cubic polynomial by \( x - 3 \). Dividing: \[\frac{2x^3 + x^2 - 18x - 9}{x - 3} = 2x^2 + 7x + 3 \]Therefore, the polynomial factors as \((x - 3)(2x^2 + 7x + 3) = 0\).

Step 6: Solve the Quadratic Equation

Now we solve \( 2x^2 + 7x + 3 = 0 \) using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 2, b = 7, c = 3 \):\[x = \frac{-7 \pm \sqrt{7^2 - 4(2)(3)}}{2(2)} = \frac{-7 \pm \sqrt{49 - 24}}{4} = \frac{-7 \pm \sqrt{25}}{4} = \frac{-7 \pm 5}{4} \]Thus, the roots are \( x = \frac{-7 + 5}{4} = -\frac{1}{2} \) and \( x = \frac{-7 - 5}{4} = -3 \).

Step 7: Compile all Solutions

The solutions to the original equation are \( x = 3, -\frac{1}{2}, -3 \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool that gives us a way to find all possible rational solutions or roots of a polynomial equation. For a polynomial of the form \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0 \), where all coefficients \( a_n, a_{n-1}, \ldots, a_0 \) are integers, the potential rational roots are expressed in the form \( \pm \frac{p}{q} \). Here, \( p \) is a factor of the constant term \( a_0 \) (in this case, \(-9\)), and \( q \) is a factor of the leading coefficient \( a_n \) (here, \(2\)).

This results in a list of possible rational roots: \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \). By testing these values in the polynomial equation, we identify any rational roots, which allow further simplification of the polynomial through factoring. It simplifies a challenging cubic equation to more manageable ones by identifying one root at a time.

Polynomial Division

Once we have found a rational root, we can use polynomial division to factor the polynomial. Polynomial division is akin to long division in arithmetic but with polynomials instead of numbers. In our example, since \( x = 3 \) is a root of the polynomial \( 2x^3 + x^2 - 18x - 9 \), we can divide the cubic polynomial by \( x - 3 \). This division helps reduce the cubic equation to a simpler quadratic one.

As a result of this division, we obtain \( 2x^2 + 7x + 3 \). This indicates that our original polynomial \( 2x^3 + x^2 - 18x - 9 \) factors into \((x - 3)(2x^2 + 7x + 3)\).

• The root used for division confirms our factorization.
• It allows us to manage complex polynomials by breaking them down into simpler equations.

Quadratic Formula

After using polynomial division, the equation simplifies to a quadratic polynomial: \( 2x^2 + 7x + 3 = 0 \). We solve this using the quadratic formula, which is a standard method for finding the roots of any quadratic equation \( ax^2 + bx + c = 0 \). The formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 7 \), and \( c = 3 \) in our example.
Substituting these values in, we calculate:
\[ x = \frac{-7 \pm \sqrt{7^2 - 4(2)(3)}}{4} = \frac{-7 \pm \sqrt{25}}{4} \]
The solutions to this quadratic equation are \( x = -\frac{1}{2} \) and \( x = -3 \).

• The quadratic formula is a straightforward, reliable method for finding the roots of any quadratic equation.
• It provides both real and complex roots.

By combining the results from these steps, we find all solutions to our original cubic equation: \( x = 3, -\frac{1}{2}, -3 \).