Question

find zeres of :
2x³-7x²-2x+1

Answer 1

2x³ - 7x² - 2x + 1

= 2x³ + x² - 8x² - 4x + 2x + 1

= x²(2x + 1) -4x (2x +1) +1(2x +1)

=(2x + 1)( x² - 4x + 1)

hence, (2x +1) and (x² - 4x + 1) are the factors of 2x³ - 7x² - 2x +1

now,
2x + 1 = 0
x = -1/2

and again,
x² - 4x + 1 = 0

use quadratic formula ,

x = { 4 ±√(16 - 4)}/2

x = { 4 ±2√3}/2 = 2±√3

hence,

-1/2 , (2 + √3) and (2-√3) are the zeros of 2x³ - 7x² -2x + 1