Solve the equation by factoring: 36 x+12x^(2)-3x^(3)=0 Answer: x=

Rearrange Equation: First, we need to rearrange the equation in standard polynomial form, which is usually written from the highest power to the lowest power of x. So, we rewrite the equation as: -3x^3 + 12x^2 + 36x = 0Factor Out GCF: Next, we can factor out the greatest common factor (GCF) from each term. The GCF in this case is 3x, so we factor it out: 3x(-x^2 + 4x + 12) = 0Factor Quadratic Expression: Now, we need to factor the quadratic expression inside the parentheses. We are looking for two numbers that multiply to -12 (the constant term) and add up to 4 (the coefficient of the x term). Those numbers are 6 and -2. So, we can write the factored form as: 3x(x - 6)(x + 2) = 0Set Equations and Solve: To find the solutions, we set each factor equal to zero and solve for x: 3x = 0 or (x - 6) = 0 or (x + 2) = 0Find Solutions: Solving each equation for x gives us the solutions: x = 0/3 or x = 6 or x = -2 Simplifying x = 0/3 gives us x = 0. So, the solutions are x = 0, x = 6, and x = -2.

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Q. Solve the equation by factoring:

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36 x+12 x^{2}-3 x^{3}=0

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Answer:

x=

Rearrange Equation: First, we need to rearrange the equation in standard polynomial form, which is usually written from the highest power to the lowest power of $x$ . So, we rewrite the equation as: $-3x^3 + 12x^2 + 36x = 0$
Factor Out GCF: Next, we can factor out the greatest common factor (GCF) from each term. The GCF in this case is $3x$ , so we factor it out: $\newline$ $3x(-x^2 + 4x + 12) = 0$
Factor Quadratic Expression: Now, we need to factor the quadratic expression inside the parentheses. We are looking for two numbers that multiply to $-12$ (the constant term) and add up to $4$ (the coefficient of the $x$ term). Those numbers are $6$ and $-2$ . So, we can write the factored form as: $3x(x - 6)(x + 2) = 0$
Set Equations and Solve: To find the solutions, we set each factor equal to zero and solve for $x$ : $3x = 0$ or $(x - 6) = 0$ or $(x + 2) = 0$
Find Solutions: Solving each equation for $x$ gives us the solutions: $\newline$ $x = \frac{0}{3}$ or $x = 6$ or $x = -2$ $\newline$ Simplifying $x = \frac{0}{3}$ gives us $x = 0$ . $\newline$ So, the solutions are $x = 0$ , $x = 6$ , and $x = -2$ .