(2−n)(6+23n)(n−2)=0

Jasonbradley

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Jun 16, 2024

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Solving the Equation: (2−n)(6+23n)(n−2) = 0

This equation represents a cubic polynomial that is already factored. To solve for the values of n that satisfy the equation, we can use the Zero Product Property. This property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero.

Let's break down the equation:

(2−n)(6+23n)(n−2) = 0

We have three factors:

• (2 - n)
• (6 + 23n)
• (n - 2)

For the entire product to equal zero, at least one of these factors must be equal to zero. Therefore, we can set each factor equal to zero and solve for n:

1. 2 - n = 0 Adding n to both sides gives us: n = 2

2. 6 + 23n = 0 Subtracting 6 from both sides and then dividing by 23 gives us: n = -6/23

3. n - 2 = 0 Adding 2 to both sides gives us: n = 2

We notice that the solution n = 2 appears twice. This indicates that the factor (n - 2) is a repeated root.

Therefore, the solutions to the equation (2−n)(6+23n)(n−2) = 0 are:

• n = 2 (with a multiplicity of 2)
• n = -6/23

This means that the equation is satisfied when n is equal to 2 (twice) or -6/23.