Problem 4 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the polynomial equation. Check your solutions. \(2 x^{4}-15 x^{3}+18 x^{2}=0\)

Step-by-Step Solution

Answer

The real solutions of the polynomial equation \(2 x^{4}-15 x^{3}+18 x^{2}=0\) are \(x=0\), \(x=3\), and \(x=3/2\).

Step 1: Factorize the polynomial

Start by taking out the common factor of \(x^2\) from the polynomial. We get \(x^2(2x^2 - 15x + 18)=0\).

Step 2: Solve the factored equation

Due to the zero-product property, a product can be zero only if at least one of the factors is zero. Thus, we solve for \(x^2=0\) and \(2x^2 - 15x + 18 = 0\) separately. For \(x^2 = 0\), the only solution is \(x=0\).

Step 3: Solve the quadratic equation

For \(2x^2 - 15x + 18 = 0\), we can use the quadratic formula \(x = [-b \pm \sqrt{(b^{2}-4ac)}]/2a\), where \(a=2\), \(b=-15\), and \(c=18\). After substituting and simplifying, we find two more solutions \(x=3\) and \(x=3/2\).

Step 4: Check your solutions

We check these solutions by substituting them back into the original equation and ensure the left hand side equals zero. Indeed, for \(x=0\), \(x=3\), and \(x=3/2\) the equation holds true.

Key Concepts

Factoring Polynomials

When solving polynomial equations, factoring is a powerful technique. It's akin to breaking down complex problems into simpler, more manageable parts. Factors are numbers or expressions you multiply together to get another number or expression. In the context of our example, the polynomial \(2x^4 - 15x^3 + 18x^2 = 0\) can be simplified by factoring out the greatest common factor first.
Here, the common factor is \(x^2\). This leads to \(x^2(2x^2 - 15x + 18) = 0\).
Factoring reduces the problem complexity and is essential for solving polynomials that are hard to solve by other means.
Understanding how to spot and pull out common factors is the first big step in simplifying and solving polynomial equations.
Additionally, if the polynomial cannot be factored easily, you might need other methods like completing the square or applying the quadratic formula.

Quadratic Formula

The quadratic formula is a universal tool for finding solutions to quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). Here, \('a'\), \('b'\), and \('c'\) represent known values and \(x\) represents the unknown variable. In general terms, the formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula is extremely useful when polynomials are tricky to factor directly.
In our example, after factoring out \(x^2\), we're left with the quadratic \(2x^2 - 15x + 18 = 0\).
Applying the quadratic formula here with \(a = 2\), \(b = -15\), and \(c = 18\) helps us find the precise roots \(x = 3\) and \(x = \frac{3}{2}\).
It’s important to understand the importance of each component in the formula:

• \(b^2 - 4ac\) is known as the discriminant, which determines the nature of the roots.
• The plus-minus symbol (\(\pm\)) indicates there may be two possible solutions.

Knowing how to apply this formula is fundamental when direct factoring isn't feasible.

Zero-Product Property

The zero-product property is a critical concept in algebra, especially when working with polynomials. It states that if the product of two or more terms equals zero, at least one of the terms must be zero. This property is what allows us to set each factor of a factored polynomial equation to zero and solve individually.
In the given polynomial \(2x^4 - 15x^3 + 18x^2 = 0\), after factoring, we have \(x^2(2x^2 - 15x + 18) = 0\).
According to the zero-product property, either \(x^2 = 0\) or \(2x^2 - 15x + 18 = 0\) must be true to satisfy the equation.
Solving these individual equations provides us with the solutions for \(x\) — in this case, the solutions are \(x = 0\), \(x = 3\), and \(x = \frac{3}{2}\).
Understanding this property not only helps in solving polynomial equations, but it also lays foundational understanding for more advanced mathematical concepts.