Problem 40
Question
Find all real solutions of the polynomial equation. $$2 x^{3}-x^{2}-18 x=-9$$
Step-by-Step Solution
Verified Answer
The solutions of the polynomial are \(x = 1\), \(x = -3/2\) and \(x = 3\)
1Step 1: Rewrite the Original Equation
Start by rewriting the equation into a form where it equals zero. This can be achieved by bringing the '-9' on the right side of the equation over to the left side of the equation. This gives a new equation: \(2 x^{3}-x^{2}-18 x + 9 = 0\)
2Step 2: Factorize the equation
The next step is to factorize the equation. Factorization in this context means expressing the equation as a product of its factors. The factorization of this equation results in \((x-1)(2x+3)(x-3)= 0\)
3Step 3: Solve for x
After factorization, the equation is now in a form where solving for 'x' can be done easily. Each factor can independently be set to zero and 'x' can be solved from each of them. From the first factor \(x-1\), if you equate it to zero and solve for 'x', x = 1. From the second factor \(2x + 3\), if you solve for 'x', x = -3/2. The third factor \(x-3\), if becomes zero for x = 3.
Key Concepts
Factoring PolynomialsSolving Polynomial EquationsReal Solutions of Equations
Factoring Polynomials
When confronted with a polynomial equation such as \(2 x^{3}-x^{2}-18 x=-9\), the first step toward finding solutions often involves factoring. Factoring polynomials is the process of breaking down a polynomial into simpler 'factor' expressions that, when multiplied together, give the original polynomial. Think of it like decomposing a number into its multiplication building blocks.
The ability to factor hinges on recognizing patterns and applying algebraic identities. Common methods include factoring out a greatest common factor, looking for pairs of terms that form a difference of squares or perfect squares, and using various factorization formulas or methods such as grouping. In our example, the polynomial was factored into \(x-1)(2x+3)(x-3) = 0\). This transformation is crucial because it simplifies finding the roots of the equation – each factor set to zero gives a potential solution.
Mastering this technique requires practice. Begin by identifying any common factors, then check for recognizable patterns that might suggest certain factorization strategies. Remember, not all polynomials are factorable with real numbers, and sometimes it might be necessary to utilize synthetic division or other advanced methods to find the factors.
The ability to factor hinges on recognizing patterns and applying algebraic identities. Common methods include factoring out a greatest common factor, looking for pairs of terms that form a difference of squares or perfect squares, and using various factorization formulas or methods such as grouping. In our example, the polynomial was factored into \(x-1)(2x+3)(x-3) = 0\). This transformation is crucial because it simplifies finding the roots of the equation – each factor set to zero gives a potential solution.
Mastering this technique requires practice. Begin by identifying any common factors, then check for recognizable patterns that might suggest certain factorization strategies. Remember, not all polynomials are factorable with real numbers, and sometimes it might be necessary to utilize synthetic division or other advanced methods to find the factors.
Solving Polynomial Equations
Once a polynomial equation is factored, as we've transformed \(2 x^{3}-x^{2}-18 x + 9 = 0\) into \(x-1)(2x+3)(x-3) = 0\), solving for \(x\) becomes a more straightforward task. This process is frequently referred to as solving polynomial equations. The Zero Product Property comes into play here, asserting that if the product of multiple factors equals zero, then at least one of the factors must equal zero.
To solve for \(x\), set each factor equal to zero and solve the resulting simple equations. In our case, we have three factors resulting in three separate equations: \(x-1=0\), \(2x+3=0\), and \(x-3=0\). Solving each gives the solutions: \(x=1\), \(x=-\frac{3}{2}\), and \(x=3\).
It's important to approach each factor as a potential source of solutions. At the end, ensure that each solution is checked back in the original polynomial to confirm its validity, as extraneous solutions may arise especially when dealing with higher degrees or complex factoring.
To solve for \(x\), set each factor equal to zero and solve the resulting simple equations. In our case, we have three factors resulting in three separate equations: \(x-1=0\), \(2x+3=0\), and \(x-3=0\). Solving each gives the solutions: \(x=1\), \(x=-\frac{3}{2}\), and \(x=3\).
It's important to approach each factor as a potential source of solutions. At the end, ensure that each solution is checked back in the original polynomial to confirm its validity, as extraneous solutions may arise especially when dealing with higher degrees or complex factoring.
Real Solutions of Equations
Finding the real solutions of an equation involves determining which values of \(x\) satisfy the original equation only under the set of real numbers. Real solutions are the points where the graph of the polynomial crosses or touches the x-axis. Calculating real solutions is paramount in various fields — from engineering to finance — because these solutions often represent feasible and tangible outcomes.
In the context of our example polynomial \(2 x^{3}-x^{2}-18 x + 9 = 0\), after factoring and solving for \(x\), we determined that the real solutions are \(x=1\), \(x=-\frac{3}{2}\), and \(x=3\). All of these are real numbers, as opposed to complex or imaginary numbers, which would have non-zero imaginary parts.
Always be mindful that equations of degree higher than two can have complex solutions as well, which won't be real numbers. For the case of cubic equations like our example, there will be at most three real solutions. Using graphing techniques or numerical methods can also help visualize or approximate the solutions, providing a practical confirmation that the solutions derived algebraically make sense graphically as well.
In the context of our example polynomial \(2 x^{3}-x^{2}-18 x + 9 = 0\), after factoring and solving for \(x\), we determined that the real solutions are \(x=1\), \(x=-\frac{3}{2}\), and \(x=3\). All of these are real numbers, as opposed to complex or imaginary numbers, which would have non-zero imaginary parts.
Always be mindful that equations of degree higher than two can have complex solutions as well, which won't be real numbers. For the case of cubic equations like our example, there will be at most three real solutions. Using graphing techniques or numerical methods can also help visualize or approximate the solutions, providing a practical confirmation that the solutions derived algebraically make sense graphically as well.
Other exercises in this chapter
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