Find the PRODUCT of the solutions to 2(x+2)^(4)-6=156 A) -5 B) 2 C) -2 D) 3

Simplify Equation: Simplify the equation to isolate the term with the exponent. We start by adding 6 to both sides of the equation to get: 2(x+2)^(4) = 162 Now, divide both sides by 2 to isolate the term with the exponent: (x+2)^(4) = 81Take Fourth Root: Take the fourth root of both sides to solve for x. Since (x+2)^(4) = 81, we can take the fourth root of both sides to get: x+2 = ±3 Because the fourth root of 81 is 3, and we consider both the positive and negative roots.Solve for x: Solve for x. We have two possible solutions for x+2: x+2 = 3 and x+2 = -3 For x+2 = 3, subtract 2 from both sides to get x = 1. For x+2 = -3, subtract 2 from both sides to get x = -5.Find Product: Find the product of the solutions. The solutions are x = 1 and x = -5. To find the product of the solutions, we multiply them together: Product = 1 * (-5) = -5

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Find the PRODUCT of the solutions to $\newline$ $2(x+2)^{4}-6=156$ $\newline$ A) $-5$ $\newline$ B) $2$ $\newline$ C) $-2$ $\newline$ D) $3$

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Algebra 2

Sum of finite series not start from 1

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Q. Find the PRODUCT of the solutions to

\newline

2(x+2)^{4}-6=156

\newline

-5

\newline

2

\newline

-2

\newline

3

Simplify Equation: Simplify the equation to isolate the term with the exponent. $\newline$ We start by adding $6$ to both sides of the equation to get: $\newline$ $2(x+2)^{4} = 162$ $\newline$ Now, divide both sides by $2$ to isolate the term with the exponent: $\newline$ $(x+2)^{4} = 81$
Take Fourth Root: Take the fourth root of both sides to solve for $x$ . $\newline$ Since $(x+2)^{4} = 81$ , we can take the fourth root of both sides to get: $\newline$ $x+2 = \pm3$ $\newline$ Because the fourth root of $81$ is $3$ , and we consider both the positive and negative roots.
Solve for x: Solve for x. $\newline$ We have two possible solutions for $x+2$ : $\newline$ $x+2 = 3$ and $x+2 = -3$ $\newline$ For $x+2 = 3$ , subtract $2$ from both sides to get $x = 1$ . $\newline$ For $x+2 = -3$ , subtract $2$ from both sides to get $x = -5$ .
Find Product: Find the product of the solutions. $\newline$ The solutions are $x = 1$ and $x = -5$ . To find the product of the solutions, we multiply them together: $\newline$ Product = $1 \times (-5) = -5$