What are the solutions to the equation 3x^4-32x^3+118x^2-152x-65=0 ?

The solutions to the equation 3x^4-32x^3+118x^2-152x-65=0 are x=5,x=-1/3 ,x=3+2i,x=3-2i

Find the zeros of 3x^4-32x^3+118x^2-152x-65=0

The zeros of 3x^4-32x^3+118x^2-152x-65=0 are x=5,x=-1/3 ,x=3+2i,x=3-2i

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3 x^{4} - 32 x^{3} + 118 x^{2} - 152 x - 65 = 0

x = 5, x = - \frac{1}{3}, x = 3 + 2 i, x = 3 - 2 i

Solution steps

3 x^{4} - 32 x^{3} + 118 x^{2} - 152 x - 65 = 0

Factor

3 x^{4} - 32 x^{3} + 118 x^{2} - 152 x - 65 : (x - 5) (3 x + 1) (x^{2} - 6 x + 13)

(x - 5) (3 x + 1) (x^{2} - 6 x + 13) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 5 = 0 or 3 x + 1 = 0 or x^{2} - 6 x + 13 = 0

Solve

x - 5 = 0 : x = 5

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

x^{2} - 6 x + 13 = 0 : x = 3 + 2 i, x = 3 - 2 i

The solutions are

x = 5, x = - \frac{1}{3}, x = 3 + 2 i, x = 3 - 2 i

x^{4} - 3 x^{3} + 2 x^{2} = 0

6 x^{3} - 11 x^{2} + x - 6 = 0

\frac{5 x ^{3}}{4 + 4} = 44

x^{5} = 0.2

x + x^{2} + x^{3} = 180

What are the solutions to the equation 3x^4-32x^3+118x^2-152x-65=0 ?
The solutions to the equation 3x^4-32x^3+118x^2-152x-65=0 are x=5,x=-1/3 ,x=3+2i,x=3-2i
Find the zeros of 3x^4-32x^3+118x^2-152x-65=0
The zeros of 3x^4-32x^3+118x^2-152x-65=0 are x=5,x=-1/3 ,x=3+2i,x=3-2i