find all the zeros of the 2x^4-9x^3+5x^2+3x-1
if two of its zeros are 2+√3 and
2-√3.
if anyone answers fast I will mark them as brainliest.
Answers
Step-by-step explanation:
Given Equation is 2x⁴ - 9x³ + 5x² + 3x - 1
Given zeroes are (2 + √3) and (2 - √3)
= {x - (2 + √3)}{x - (2 - √3)}
= (x - 2 - √3)(x - 2 + √3)
= (x - 2)² - (√3)²
= x² + 4 - 4x - 3
= x² - 4x + 1
Divide f(x) by x² - 4x + 1.
Long Division Method:
x² - 4x + 1) 2x⁴ - 9x³ + 5x² + 3x - 1(2x² - x - 1
2x⁴ - 8x³ + 2x²
---------------------------------
-x³ + 3x² + 3x
-x³ + 4x² - x
-----------------------------------
-x² + 4x - 1
-x² + 4x - 1
--------------------------------
0
Dividend = Divisor * Quotient + Remainder
= (x² - 4x + 1)(2x² - x - 1)
= (x² - 4x - 1)(2x² - 2x + x - 1)
= (x² - 4x - 1)(2x(x - 1) + 1(x - 1))
= (x² - 4x + 1)(2x + 1)(x - 1)
= (x - 2 - √3)(x - 2 + √3)(2x + 1)(x - 1)
Therefore, Zeroes are : 2 ± √3, -1/2, 1
Hope it helps!