Problem 19
Question
For the following exercises, solve the following polynomial equations by grouping and factoring. $$ 5 x^{3}+45 x=2 x^{2}+18 $$
Step-by-Step Solution
Verified Answer
The solution is \( x = \frac{2}{5} \).
1Step 1: Move All Terms to One Side
Subtract the terms on the right side of the equation from both sides to set the equation to zero:\[ 5x^3 + 45x - 2x^2 - 18 = 0 \]
2Step 2: Rearrange the Terms
Rearrange the terms in descending order of powers of \( x \): \[ 5x^3 - 2x^2 + 45x - 18 = 0 \]
3Step 3: Group Terms and Factor Common Factors
Group terms to factor by grouping. Group the first two and the last two terms:\[ (5x^3 - 2x^2) + (45x - 18) = 0 \]Factor out common factors from each group:\[ x^2(5x - 2) + 9(5x - 2) = 0 \]
4Step 4: Factor Out the Common Binomial
Notice that \((5x - 2)\) is a common factor in both groups. Factor \((5x - 2)\) out:\[ (x^2 + 9)(5x - 2) = 0 \]
5Step 5: Solve Each Factor
Set each factor to zero and solve for \( x \):For \( x^2 + 9 = 0 \):\[ x^2 = -9 \]This gives no real solutions since negative numbers do not have real square roots.For \( 5x - 2 = 0 \):\[ 5x = 2 \]\[ x = \frac{2}{5} \]
6Step 6: Conclusion
The only real solution to the polynomial equation is \( x = \frac{2}{5} \).
Key Concepts
FactoringGroupingReal Solutions
Factoring
Factoring is one of the primary techniques used in solving polynomial equations. It involves expressing a polynomial as a product of its factors, which are simpler polynomials, often linear or quadratic in nature. This approach breaks down complex equations into more manageable parts, allowing us to leverage properties like the zero-product property. The zero-product property states that if the product of several factors is zero, at least one of the factors must be zero.
In our given exercise, the equation was re-arranged so we could apply factoring effectively. First, terms on one side of the equation were collected to equal zero, then arranged in decreasing powers of the variable, which is a common strategy to simplify the equation.
Factoring involves careful observation to identify common factors among terms, as seen in the grouping technique used here. Recognizing patterns, such as common binomials that can be factored out, further simplifies solving polynomials. Developing a keen eye for these factors allows you to solve polynomial equations efficiently and correctly.
In our given exercise, the equation was re-arranged so we could apply factoring effectively. First, terms on one side of the equation were collected to equal zero, then arranged in decreasing powers of the variable, which is a common strategy to simplify the equation.
Factoring involves careful observation to identify common factors among terms, as seen in the grouping technique used here. Recognizing patterns, such as common binomials that can be factored out, further simplifies solving polynomials. Developing a keen eye for these factors allows you to solve polynomial equations efficiently and correctly.
Grouping
Grouping is a strategic approach for factoring that involves organizing terms with similar characteristics together. This technique is particularly useful for polynomials where straightforward factoring is not immediately possible. By grouping terms, you prepare the polynomial for factorization by identifying and isolating common factors.
In practice, as seen in the original exercise, terms of the polynomial are grouped into pairs. For instance, in the equation \[ 5x^3 - 2x^2 + 45x - 18 = 0 \] terms were grouped as follows:
In our example, this method made it easier to spot the common binomial, \((5x - 2)\), allowing us to reduce the polynomial into a more solvable form. Grouping is a valuable skill, turning a complicated polynomial into a sequence of simpler tasks.
In practice, as seen in the original exercise, terms of the polynomial are grouped into pairs. For instance, in the equation \[ 5x^3 - 2x^2 + 45x - 18 = 0 \] terms were grouped as follows:
- First Group: \(5x^3 - 2x^2\)
- Second Group: \(45x - 18\)
In our example, this method made it easier to spot the common binomial, \((5x - 2)\), allowing us to reduce the polynomial into a more solvable form. Grouping is a valuable skill, turning a complicated polynomial into a sequence of simpler tasks.
Real Solutions
Real solutions to polynomial equations are the solutions that are observable within the set of real numbers. These are the solutions without imaginary components and can be plotted on the real number line.
In the original equation \[ (x^2 + 9)(5x - 2) = 0 \] our task was to find the real solutions. By setting each factor equal to zero, \[ x^2 + 9 = 0 \] and \[ 5x - 2 = 0 \] we proceed with solving for \(x\):
In the original equation \[ (x^2 + 9)(5x - 2) = 0 \] our task was to find the real solutions. By setting each factor equal to zero, \[ x^2 + 9 = 0 \] and \[ 5x - 2 = 0 \] we proceed with solving for \(x\):
- \(x^2 + 9 = 0\) resulted in no real solutions because square roots of negative numbers are not real.
- \(5x - 2 = 0\) gives a real solution of \(x = \frac{2}{5}\).
Other exercises in this chapter
Problem 18
Solve each rational equation for x. State all x-values that are excluded from the solution set. \(\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{(x-1)(x-2)}\)
View solution Problem 19
For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation. $$ |x-2|+4 \geq 10 $$
View solution Problem 19
Solve the inequality involving absolute value. Write your final answer in interval notation. $$ |x-2|+4 \geq 10 $$
View solution Problem 19
For the following exercises, perform the indicated operation and express the result as a simplified complex number. $$ (-4+4 i)-(-6+9 i) $$
View solution