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Find all the zeroes of the equation: x^4 – 6x^2 – 7x – 6 = 0?

I need to find both real and imaginary. I have trouble breaking down the problem in order to solve. Can anyone show me the steps and the answer?!

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The Best Answer for Find all the zeroes of the equation: x^4 – 6x^2 – 7x – 6 = 0?

Let f(x) = x^4 – 6x^2 – 7x – 6

Note factors of the constant term –6 are ±1, ±2, ±3 and ±6

f(−2) = 0 means x+2 is a factor of f(x)

f(3) = 0 means x−3 is a factor of f(x)

Do long division or synthetic division or by inspection

f(x) = (x+2)(x−3)(x²+x+1)

(x+2)(x−3)(x²+x+1) = 0

x = −2 or x = 3 or x²+x+1 = 0

Solving the quadratic equation, in addition to the 2 real solutions,

x = ½(−1 ± i√3).

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Zeros are found from the factors

x^4 - 6x^2 -7x -6 = 0 = (x-3)(x+2)(x^2 + x + 1); the first two factors give x = -2 and 3, the 3rd has the imaginary roots when x^2+x+1 = 0 which the quadratic formula gives as x = (-1/2) +/- [sqrt(-3)]/2

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