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The solutions to the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0 are x = 0, x = -1/2, and x = 3.
To solve the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0, we can start by factoring out the common factor, which is x^3:
x^3(2x^3 - 3x^2 - 7x - 6) = 0
Now, we can set each factor equal to zero:
x^3 = 0
2x^3 - 3x^2 - 7x - 6 = 0
Solving the first equation, we find that x = 0 is a solution.
For the second equation, we can use factoring or the quadratic formula to find the remaining solutions. Factoring the quadratic equation, we get:
(2x + 1)(x^2 - 2x - 6) = 0
Setting each factor equal to zero, we have:
2x + 1 = 0
x^2 - 2x - 6 = 0
Solving these equations, we find that x = -1/2 and x = 3 are the remaining solutions.
Therefore, the solutions to the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0 are x = 0, x = -1/2, and x = 3.
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Rewritten by : Barada
The solutions to the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0 are x = 0, x = -1/2, and x = 3.
To solve the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0, we can start by factoring out the common factor, which is x^3:
x^3(2x^3 - 3x^2 - 7x - 6) = 0
Now, we can set each factor equal to zero:
x^3 = 0
2x^3 - 3x^2 - 7x - 6 = 0
Solving the first equation, we find that x = 0 is a solution.
For the second equation, we can use factoring or the quadratic formula to find the remaining solutions. Factoring the quadratic equation, we get:
(2x + 1)(x^2 - 2x - 6) = 0
Setting each factor equal to zero, we have:
2x + 1 = 0
x^2 - 2x - 6 = 0
Solving these equations, we find that x = -1/2 and x = 3 are the remaining solutions.
Therefore, the solutions to the equation 2x^6 - 3x^5 - 7x^4 - 6x^3 = 0 are x = 0, x = -1/2, and x = 3.
Learn more about solving polynomial equations here:
https://brainly.com/question/14837418
#SPJ14
