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Find all the real zeros of the function f(x)=2x^(3)-19x^(2)+38x+24 given that x-4 is a factor.

User DooBLER
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To find all the real zeros of the polynomial f(x) = 2x^3 - 19x^2 + 38x + 24, first, knowing that x-4 is a factor, we can divide the polynomial by x-4. The result of the division is a quadratic equation, because when you divide a cubic polynomial by a linear factor, you get a quadratic equation.

First, let's simplify the cubic polynomial f(x) = 2x^3 - 19x^2 + 38x + 24 by the factor (x - 4).

The simplified form is another polynomial, let's call it q(x). So, f(x) = (x - 4) * q(x).

Let's use polynomial long division or Synthetic Division to divide the cubic polynomial by (x - 4) and find q(x).

The result is q(x) = 2x^2 - 3x - 6.

Next, we will find the zeros of this quadratic equation q(x) = 2x^2 - 3x - 6.

To find the zeros, set the equation equal to zero and solve for x:

2x^2 - 3x - 6 = 0

You can solve it using the quadratic formula:

x = [-(-3) ± sqrt((-3)^2 - 4*2*(-6))] / (2*2)
x = (3 ± sqrt(9 + 48)) / 4
x = (3 ± sqrt(57)) / 4
x = 3/4 ± sqrt(57)/4

Therefore, the roots of the cubic function f(x) = 2x^3 - 19x^2 + 38x + 24 are x = 4, 3/4 - sqrt(57)/4, 3/4 + sqrt(57)/4.

User Dewsworld
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