Solve for x x^3-5x^2=-6x
The problem asks for determining the value(s) of the variable x that satisfies the equation x^3 - 5x^2 + 6x = 0. It is essentially a cubic equation where one needs to manipulate algebraically or use various solving techniques to find the roots of the equation. The question is looking for a numerical solution or solutions that make the equation true.
$x^{3} - 5 x^{2} = - 6 x$
Answer
Solution:
Step 1
Move $6x$ to the left side of the original equation to obtain a quadratic equation.
$$x^3 - 5x^2 + 6x = 0$$
Step 2
Begin factoring the polynomial on the left side.
Step 2.1
Extract the common factor $x$ from each term.
$$x(x^2 - 5x + 6) = 0$$
Step 2.2
Further factor the quadratic expression within the parentheses.
Step 2.2.1
Apply the AC method to factor $x^2 - 5x + 6$.
Find two numbers that multiply to $6$ (the constant term) and add up to $-5$ (the coefficient of $x$).
The numbers are $-3$ and $-2$.
Step 2.2.2
Rewrite the quadratic expression as a product of two binomials using the numbers found.
$$x((x - 3)(x - 2)) = 0$$
Step 3
Recognize that for the product to be zero, at least one of the factors must be zero.
Step 4
Set the first factor equal to zero and solve for $x$.
$$x = 0$$
Step 5
Set the second factor equal to zero and solve for $x$.
$$x - 3 = 0 \Rightarrow x = 3$$
Step 6
Set the third factor equal to zero and solve for $x$.
$$x - 2 = 0 \Rightarrow x = 2$$
Step 7
Combine all the solutions to find the complete set of values for $x$ that satisfy the equation.
$$x = 0, 3, 2$$
Knowledge Notes:
To solve the cubic equation $x^3 - 5x^2 = -6x$, we first move all terms to one side to form a polynomial set equal to zero. This allows us to use factoring techniques to find the solutions for $x$.
The factoring process involves finding common factors and using methods such as the AC method to factor quadratic expressions. The AC method is particularly useful for factoring trinomials of the form $ax^2 + bx + c$ where $a = 1$. It involves finding two numbers that multiply to the constant term $c$ and add up to the coefficient $b$.
Once the polynomial is factored, we use the Zero Product Property, which states that if the product of several factors is zero, then at least one of the factors must be zero. Setting each factor equal to zero and solving for $x$ gives us the roots of the polynomial.
In this problem, the roots are $x = 0, 3, 2$. These are the solutions to the original cubic equation.
