Problem 54
Question
$$ m^{3}+8 m^{2}+15 m=0 $$
Step-by-Step Solution
Verified Answer
The solutions are: $$m = 0, -3, -5.$$
1Step 1 - Identify the Polynomial
Recognize the given equation is a polynomial in the form of a cubic function: $$m^{3} + 8m^{2} + 15m = 0.$$
2Step 2 - Factor Out the Greatest Common Factor (GCF)
First, factor out the common term, which is 'm' in this case: $$m (m^{2} + 8 m + 15) = 0.$$
3Step 3 - Factor the Quadratic
Next, factor the quadratic expression inside the parentheses. We need to find two numbers that multiply to 15 and add to 8: $$m (m + 3)(m + 5) = 0.$$
4Step 4 - Set Each Factor to Zero
Set each factor equal to zero to find the solutions for 'm': $$m = 0$$ $$m + 3 = 0$$ $$m + 5 = 0.$$
5Step 5 - Solve for 'm'
Solve the equations from Step 4: $$m = 0$$ $$m = -3$$ $$m = -5.$$
6Step 6: Conclusion
Combine the solutions to state the final answer: The solutions are: $$m = 0, -3, -5.$$
Key Concepts
Cubic FunctionsGreatest Common FactorQuadratic EquationsSolving Equations
Cubic Functions
Cubic functions are polynomial functions of degree three. The general form looks like this:
\(ax^3 + bx^2 + cx + d = 0\).
They are called 'cubic' because the highest power of the variable (commonly 'x') is three.
In our exercise, the equation \(m^3 + 8m^2 + 15m = 0\) is a cubic function.
Cubic functions can have one, two, or three real roots (solutions).
These are the points where the function intersects the x-axis.
\(ax^3 + bx^2 + cx + d = 0\).
They are called 'cubic' because the highest power of the variable (commonly 'x') is three.
In our exercise, the equation \(m^3 + 8m^2 + 15m = 0\) is a cubic function.
Cubic functions can have one, two, or three real roots (solutions).
These are the points where the function intersects the x-axis.
Greatest Common Factor
To make solving polynomial equations easier, you often start by factoring out the Greatest Common Factor (GCF).
The GCF of a polynomial is the largest term (factor) that each term in the polynomial shares.
In the equation \(m^3 + 8m^2 + 15m = 0\), we see that each term has at least one 'm'.
So, we can factor 'm' out: \(m(m^2 + 8m + 15) = 0\).
This step simplifies the equation and reveals a quadratic equation in the parentheses.
The GCF of a polynomial is the largest term (factor) that each term in the polynomial shares.
In the equation \(m^3 + 8m^2 + 15m = 0\), we see that each term has at least one 'm'.
So, we can factor 'm' out: \(m(m^2 + 8m + 15) = 0\).
This step simplifies the equation and reveals a quadratic equation in the parentheses.
Quadratic Equations
A quadratic equation is a polynomial of degree two, and its general form is \(ax^2 + bx + c = 0\).
In the original exercise, the quadratic equation inside the parentheses is \(m^2 + 8m + 15\).
To factor this quadratic equation, look for two numbers that multiply to 15 (the constant term) and add up to 8 (the coefficient of the linear term):
These numbers are 3 and 5 because \(3 * 5 = 15\) and \(3 + 5 = 8\).
Thus, the factored form is \((m + 3)(m + 5)\).
So, \(m(m^2 + 8m + 15)\) becomes \(m(m + 3)(m + 5) = 0\).
In the original exercise, the quadratic equation inside the parentheses is \(m^2 + 8m + 15\).
To factor this quadratic equation, look for two numbers that multiply to 15 (the constant term) and add up to 8 (the coefficient of the linear term):
These numbers are 3 and 5 because \(3 * 5 = 15\) and \(3 + 5 = 8\).
Thus, the factored form is \((m + 3)(m + 5)\).
So, \(m(m^2 + 8m + 15)\) becomes \(m(m + 3)(m + 5) = 0\).
Solving Equations
After factoring the polynomial equation, the next step is solving for 'm'.
You set each factor equal to zero:
Solving these equations gives us the solutions:
This means that these values are the roots of the cubic function, or the points where the function intersects the x-axis.
You set each factor equal to zero:
- \(m = 0\)
- \(m + 3 = 0\)
- \(m + 5 = 0\)
Solving these equations gives us the solutions:
- \(m = 0\)
- \(m = -3\)
- \(m = -5\)
This means that these values are the roots of the cubic function, or the points where the function intersects the x-axis.
Other exercises in this chapter
Problem 53
For exercises 53-58, use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ y^{4}-4 y^{2}-32 $$
View solution Problem 53
(a) factor by grouping. Identify any prime polynomials. (b) check. $$ 3 m^{2}-3 m v-m-v $$
View solution Problem 54
Factor completely. Identify any prime polynomials. $$ 4 p^{2} z-20 p^{2}+p w z-5 p w $$
View solution Problem 54
Use a pattern to factor. Check. Identify any prime polynomials. $$ 64 c^{2}+48 c+9 $$
View solution