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The function f is defined by f(x) = 2x³ + 3x² + cx + 8, where c is a constant. In the xy-plane, the graph of f intersects the x-axis at the three points (−4, 0), (1/2, 0), and (p, 0). What is the value of c?​

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User Elias N
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1 Answer

5 votes

Answer:

  • A. -18

Step-by-step explanation:

Given the function
\tt{f(x) = 2x^3 + 3x^2 + cx + 8} and the x-intercepts
\tt{(-4, 0)} and
\tt{((1)/(2), 0)}, we can find the value of c.

To find the value of c, we can use the fact that the product of the roots of a cubic equation is equal to the constant term divided by the coefficient of the highest power of
\bf{x}.

In this case:

  • the constant term = 8
  • the coefficient of the highest power of x = 2

Using the given x-intercepts, we have:

1.
\tt{(-4, 0)}: This means that when
\tt{x = -4}, the function
\tt{f(x)} equals 0.

Plugging this into the equation, we get:


  • \tt{f(-4) = 2(-4)^3 + 3(-4)^2 + c(-4) + 8 = 0}

Simplifying this equation, we get:


  • \tt{-128 + 48 - 4c + 8 = 0}

  • \tt{-72 - 4c = 0}

  • \tt{-4c = 72}

  • \tt{c = -18}

Therefore, the value of c is -18.

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Full Question

The function f is defined by f(x) = 2x³ + 3x² + cx + 8, where c is a constant. In-example-1
User Takepara
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