What are the solutions to the equation 2x^5+3x^4-5x^3-5x^2+3x+2=0 ?

The solutions to the equation 2x^5+3x^4-5x^3-5x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=-2

Find the zeros of 2x^5+3x^4-5x^3-5x^2+3x+2=0

The zeros of 2x^5+3x^4-5x^3-5x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{5} + 3 x^{4} - 5 x^{3} - 5 x^{2} + 3 x + 2 = 0

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

Decimal Notation

x = - 1, x = 1, x = - 0.5, x = - 2

Solution steps

2 x^{5} + 3 x^{4} - 5 x^{3} - 5 x^{2} + 3 x + 2 = 0

Factor

2 x^{5} + 3 x^{4} - 5 x^{3} - 5 x^{2} + 3 x + 2 : (x + 1) (x - 1)^{2} (2 x + 1) (x + 2)

(x + 1) (x - 1)^{2} (2 x + 1) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x + 2 = 0 : x = - 2

The solutions are

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

m^{4} = 1

(0.2 + 2 x)^{2} (x) = 0.000015

(x - 1) (x - 2) (x - 3) (x - 4) = 15

a, (16)^{3} = a^{3} a^{2} b b^{2} a b^{3}

d^{2} j^{5} x^{4} - 22 x^{2} + 8 = 2

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