What are the solutions to the equation 6x^5+11x^4-39x^3-41x^2+45x+18=0 ?

The solutions to the equation 6x^5+11x^4-39x^3-41x^2+45x+18=0 are x=1,x=2,x=-3,x=-1/3 ,x=-3/2

Find the zeros of 6x^5+11x^4-39x^3-41x^2+45x+18=0

The zeros of 6x^5+11x^4-39x^3-41x^2+45x+18=0 are x=1,x=2,x=-3,x=-1/3 ,x=-3/2

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6 x^{5} + 11 x^{4} - 39 x^{3} - 41 x^{2} + 45 x + 18 = 0

x = 1, x = 2, x = - 3, x = - \frac{1}{3}, x = - \frac{3}{2}

Decimal Notation

x = 1, x = 2, x = - 3, x = - 0.33333 \dots, x = - 1.5

Solution steps

6 x^{5} + 11 x^{4} - 39 x^{3} - 41 x^{2} + 45 x + 18 = 0

Factor

6 x^{5} + 11 x^{4} - 39 x^{3} - 41 x^{2} + 45 x + 18 : (x - 1) (x - 2) (x + 3) (3 x + 1) (2 x + 3)

(x - 1) (x - 2) (x + 3) (3 x + 1) (2 x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 2 = 0 or x + 3 = 0 or 3 x + 1 = 0 or 2 x + 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

Solve

x + 3 = 0 : x = - 3

Solve

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

Solve

2 x + 3 = 0 : x = - \frac{3}{2}

The solutions are

x = 1, x = 2, x = - 3, x = - \frac{1}{3}, x = - \frac{3}{2}

5 x^{3} = 320

x^{3} - 5 x^{2} - 4 x + 20

x^{3} - 2 x^{2} + x - 2

x, (x^{2} + y)^{3} = y

x^{3} - 9 x^{2} + 33 x - 65 = 0

What are the solutions to the equation 6x^5+11x^4-39x^3-41x^2+45x+18=0 ?
The solutions to the equation 6x^5+11x^4-39x^3-41x^2+45x+18=0 are x=1,x=2,x=-3,x=-1/3 ,x=-3/2
Find the zeros of 6x^5+11x^4-39x^3-41x^2+45x+18=0
The zeros of 6x^5+11x^4-39x^3-41x^2+45x+18=0 are x=1,x=2,x=-3,x=-1/3 ,x=-3/2