6x^4+4x^2-3x+2=2x^2(3x^2-1)

Question

6x^4+4x^2-3x+2=2x^2(3x^2-1)

Answer 1

6x⁴ + 4x² - 3x + 2 = 2x²(3x² - 1)

 

 

We can distribute the terms on the right side of equation.

 

6x⁴ + 4x² - 3x + 2 = 6x⁴ - 2x²

 

 

Now we can make the right side equal to zero by subtract 6x⁴ and adding 2x² on both sides of the equation.

 

6x² - 3x + 2 = 0

 

 

Use the quadratic formula to solve for x:

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 6

b = -3

c = 2

 

Plug in these values into the formula.  Because of the plus/minus sign in the formula, there will be two values of x.