Question

Solve the following equations: 2x^3+3x^2+3x+1=0

Answer 1

Answer:

Given equation, 2x

3

+3x

2

+3x+1=0

⟹2x

3

+2x

2

+x

2

+2x+x+1=0⟹(2x+1)(x

2

+x+1)=0

Solving the 2 equations, we have x=

2

−1

,

2

−1±

3

i

Answer 2

Step-by-step explanation:

$2x ^ { 3 } +3x ^ { 2 } +3x+1=0$

$±\frac{1}{2},±1$

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

$By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+3x^{2}+3x+1 by 2\left(x+\frac{1}{2}\right)=2x+1 to get x^{2}+x+1. Solve the equation where the result equals to 0.$

$x^{2}+x+1=0$

$All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 1 for b, and 1 for c in the quadratic formula.$

$x=\frac{-1±\sqrt{1^{2}-4\times 1\times 1}}{2}$

$x=\frac{-1±\sqrt{-3}}{2}$

$Since the square root of a negative number is not defined in the real field, there are no solutions.$

$x\in \emptyset$

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