algebra 2 problem

Question

I need help solving this equation in the real number system

 

3x⁴ - 16x³ + 29x² - 20x + 4 = 0 what is x in this case

Answer 1

Looking at the leading and constant coefficients of 3x⁴ - 16x³ + 29x² - 20x + 4 = 0.

 

By the rational root test, every rational root A/B if it exist has A be a factor of the constant coefficient and B be a factor of the leading one.

 

In this case A ∈ {±1,±2,±4} and B ∈ {±1,±3}

 

So the possible rational roots are ±1, ±2, ±4, ±1/3, ±2/3, ±4/3.

 

Three candidates are roots, x=1/3, x=1, and x=2.

 

Note that these are all the roots and one of them is a double root. To see which one, factor the polynomial

 

The roots imply that 3x-1, x-1, and x-2 are factors. Using polynomial long division repeatedly, you will get

 

3x⁴ - 16x³ + 29x² - 20x + 4 = (3x - 1)(x - 1)(x - 2)².

 

Therefore x = 2 is the double root.