Understand Relationship Between Factorials: We are given the equation ((x!)/(x-2)!) = 12. To solve for x, we need to understand the relationship between x! and (x-2)!.Rewrite and Simplify Expression: The factorial of a number n, denoted as n!, is the product of all positive integers less than or equal to n. Therefore, x! = x * (x-1) * (x-2) * ... * 1.Solve Quadratic Equation: Similarly, (x-2)! = (x-2) * (x-3) * ... * 1. Notice that (x-2)! is a part of the product in x!.Factor and Solve for x: We can rewrite x! as x * (x-1) * (x-2)!. Now, we can divide x! by (x-2)! to simplify the expression.Check Validity and Final Solution: Dividing x! by (x-2)! gives us (x * (x-1) * (x-2)!)/(x-2)! which simplifies to x * (x-1) because the (x-2)! terms cancel out.Now we have the equation x * (x-1) = 12. To find the value of x, we need to solve this quadratic equation.We can expand the equation to x^2 - x - 12 = 0 to solve for x.To solve the quadratic equation x^2 - x - 12 = 0, we can factor it. The factors of -12 that add up to -1 are -4 and 3.Factoring the quadratic equation gives us (x - 4)(x + 3) = 0.Setting each factor equal to zero gives us two possible solutions for x: x - 4 = 0 or x + 3 = 0.Solving x - 4 = 0 gives us x = 4. Solving x + 3 = 0 gives us x = -3.However, since we are dealing with factorials, x must be a non-negative integer. Therefore, x = -3 is not a valid solution.The only valid solution is x = 4. We can check this by substituting x = 4 into the original equation to see if it equals 12.Substituting x = 4 into the original equation gives us ((4!)/(4-2)!) = (24/2) = 12, which is true.

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$\frac{x!}{(x-2)!}=12$

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Algebra 2

Add, subtract, multiply, and divide polynomials

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\frac{x!}{(x-2)!}=12

Understand Relationship Between Factorials: We are given the equation $\frac{x!}{(x-2)!} = 12$ . To solve for $x$ , we need to understand the relationship between $x!$ and $(x-2)!$ .
Rewrite and Simplify Expression: The factorial of a number $n$ , denoted as $n!$ , is the product of all positive integers less than or equal to $n$ . Therefore, $x! = x \times (x-1) \times (x-2) \times \ldots \times 1$ .
Solve Quadratic Equation: Similarly, $(x-2)! = (x-2) \times (x-3) \times \ldots \times 1$ . Notice that $(x-2)!$ is a part of the product in $x!$ .
Factor and Solve for $x$ : We can rewrite $x!$ as $x \times (x-1) \times (x-2)!$ . Now, we can divide $x!$ by $(x-2)!$ to simplify the expression.
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .Setting each factor equal to zero gives us two possible solutions for $x$ : $(x-2)!$ $6$ or $(x-2)!$ $7$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x * (x-1) * (x-2)!)/(x-2)!$ which simplifies to $x * (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x * (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .Setting each factor equal to zero gives us two possible solutions for $x$ : $(x-2)!$ $6$ or $(x-2)!$ $7$ .Solving $(x-2)!$ $6$ gives us $(x-2)!$ $9$ . Solving $(x-2)!$ $7$ gives us $(x * (x-1) * (x-2)!)/(x-2)!$ $1$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .Setting each factor equal to zero gives us two possible solutions for $x$ : $(x-2)!$ $6$ or $(x-2)!$ $7$ .Solving $(x-2)!$ $6$ gives us $(x-2)!$ $9$ . Solving $(x-2)!$ $7$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ .However, since we are dealing with factorials, $x$ must be a non-negative integer. Therefore, $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ is not a valid solution.
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .Setting each factor equal to zero gives us two possible solutions for $x$ : $(x-2)!$ $6$ or $(x-2)!$ $7$ .Solving $(x-2)!$ $6$ gives us $(x-2)!$ $9$ . Solving $(x-2)!$ $7$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ .However, since we are dealing with factorials, $x$ must be a non-negative integer. Therefore, $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ is not a valid solution.The only valid solution is $(x-2)!$ $9$ . We can check this by substituting $(x-2)!$ $9$ into the original equation to see if it equals $(x \times (x-1) \times (x-2)!)/(x-2)!$ $6$ .
Check Validity and Final Solution: Dividing $x!$ by $(x-2)!$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ which simplifies to $x \times (x-1)$ because the $(x-2)!$ terms cancel out.Now we have the equation $x \times (x-1) = 12$ . To find the value of $x$ , we need to solve this quadratic equation.We can expand the equation to $x^2 - x - 12 = 0$ to solve for $x$ .To solve the quadratic equation $x^2 - x - 12 = 0$ , we can factor it. The factors of $(x-2)!$ $0$ that add up to $(x-2)!$ $1$ are $(x-2)!$ $2$ and $(x-2)!$ $3$ .Factoring the quadratic equation gives us $(x-2)!$ $4$ .Setting each factor equal to zero gives us two possible solutions for $x$ : $(x-2)!$ $6$ or $(x-2)!$ $7$ .Solving $(x-2)!$ $6$ gives us $(x-2)!$ $9$ . Solving $(x-2)!$ $7$ gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ .However, since we are dealing with factorials, $x$ must be a non-negative integer. Therefore, $(x \times (x-1) \times (x-2)!)/(x-2)!$ $1$ is not a valid solution.The only valid solution is $(x-2)!$ $9$ . We can check this by substituting $(x-2)!$ $9$ into the original equation to see if it equals $(x \times (x-1) \times (x-2)!)/(x-2)!$ $6$ .Substituting $(x-2)!$ $9$ into the original equation gives us $(x \times (x-1) \times (x-2)!)/(x-2)!$ $8$ , which is true.

$\frac{x!}{(x-2)!}=12$

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x!(x−2)!=12\frac{x!}{(x-2)!}=12(x−2)!x!​=12

$\frac{x!}{(x-2)!}=12$