Problem 100 - Elementary Algebra

Question

For exercises 97-100, the completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Solve: \(x^{3}+7 x^{2}+10 x=0\) Incorrect Answer: \(x^{3}+7 x^{2}+10 x=0\) \(x\left(x^{2}+7 x+10\right)=0\) \((x+5)(x+2)=0\) \(x+5=0\) or \(x+2=0\) \(\frac{-5-5}{x=-5}\) or \(\frac{-2-2}{x=-2}\) Solutions: \(x=-5, x=-2\)

Step-by-Step Solution

Answer

The solutions are: \( x=0, x=-2, x=-5 \)

Step 1: Identify the Mistake

The problem is solved incorrectly in the factoring step and solving for the roots. The correct factored form and solving for x have mistakes.

Step 2: Rewrite the Problem

Rewrite the given problem correctly. Solve: \[ x^{3}+7x^{2}+10x=0 \]

Step 3: Factor Out the Common Term

Factor out the common term from the equation. \[ x(x^{2}+7x+10)=0 \]

Step 4: Further Factor the Quadratic Expression

Further factor the quadratic expression inside the parenthesis. \[ x(x+2)(x+5)=0 \]

Step 5: Set Each Factor to Zero

Set each factor of the equation equal to zero and solve for x. \[ x=0,\ (x+2)=0,\ (x+5)=0 \]

Step 6: Solve for x

Solving for x, we get the solutions: \[ x=0 \] \[ x=-2 \] \[ x=-5 \]

Step 7: Final Step: Write the Correct Solution

The correct roots of the equation are: \[ x=0,\ x=-2,\ x=-5 \]

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential step in solving many algebraic equations, including cubic equations. When you factor a polynomial, you're breaking it down into simpler components called factors. These factors, when multiplied together, give you the original polynomial. For example, in the equation \(x^{3}+7x^{2}+10x=0\), the common term is \(x\). So the first step is to factor out \(x\), resulting in \(x(x^{2}+7x+10)=0\).

Once you have factored out the common term, the next step is to address the quadratic equation left inside the parenthesis. You look for two numbers that multiply to give you the constant term (10) and add up to give you the coefficient of the middle term (7). In this case, the numbers are 2 and 5. So, \(x^{2}+7x+10\) can be further factored to \((x+2)(x+5)\).

Therefore, the fully factored form of the original cubic equation is \(x(x+2)(x+5)=0\). Successfully factoring the polynomial helps to simplify the process of solving for the roots.

Setting Factors to Zero

Once you have factored the polynomial completely, the next step is to set each factor equal to zero. This comes from the zero-product property, which states that if a product of several factors equals zero, at least one of the factors must be zero. For the cubic equation \(x(x+2)(x+5)=0\), setting each factor to zero gives:

• \(x=0\)
• \(x+2=0\)
• \(x+5=0\)

By setting each factor to zero, we effectively break the problem down into simpler, solvable parts. This is crucial for isolating each possible solution for the variable \(x\).

Solving for Roots

The final step involves solving for the variable \(x\) from the equations derived from setting each factor to zero. These equations are straightforward and involve simple algebraic manipulation.

To solve \(x=0\), it is already in its simplest form, so the solution is \(x=0\).

For \(x+2=0\), subtract 2 from both sides to isolate \(x\):
\[ x+2-2 = 0-2 \]
\[ x = -2 \]

For \(x+5=0\), subtract 5 from both sides:
\[ x+5-5 = 0-5 \]
\[ x = -5 \]

Therefore, the cubic equation \(x^{3}+7x^{2}+10x=0\) has three solutions: \(x=0\), \(x=-2\), and \(x=-5\). Understanding how to solve for the roots after setting factors to zero is key to finding all possible solutions to the equation.