Find all the zeroes of 2x4 -6x3 +3x2 +3x -2 ,if two of zeroes are 1/root2 and -1/ root 2?

Question

Answer

Let a = 1√2 nd B= -1√2 a+B = 0 a.B = -3 a polynomial whOse zero a nd B =)x2- (a+B)x+ aB =)x2+(-3) =) x2 - 3x2-3 ) 2x4- 6x3 +3x2 +3x-2 ( divided this eqN u got ur answer••••••

Answer

Answer

Introduction: In this problem, we are given a polynomial equation and two of its zeroes. We need to find all the zeroes of the equation.

Given:
Equation: 2x⁴ - 6x³ + 3x² + 3x - 2
Two zeroes: 1/√2 and -1/√2

Solution:

To find all the zeroes of the equation, we can use the factor theorem and synthetic division.

Step 1: Use the given zeroes to get two factors of the equation.
(x - 1/√2) and (x + 1/√2)

Step 2: Use synthetic division to divide the equation by these factors.

For (x - 1/√2):
1/√2 | 2 -6 3 3 -2
| √2 -2 -√2 1
------------------------
2 -6+2√2 1-√2 4-√2

So, (x - 1/√2) is a factor of 2x⁴ - 6x³ + 3x² + 3x - 2 and the quotient is 2x³ - 6x² + 1x - (4-√2)

For (x + 1/√2):
-1/√2 | 2 -6+2√2 1-√2 4-√2
| -√2 -2 √2
------------------------
2 -6 -1 2-√2

So, (x + 1/√2) is a factor of 2x⁴ - 6x³ + 3x² + 3x - 2 and the quotient is 2x³ - 6x² - 1x + (2-√2)

Step 3: Factor the quotient obtained in step 2 using grouping.

2x³ - 6x² + 1x - (4-√2) = (2x³ - 6x²) + (1x - (4-√2))
= 2x²(x - 3) + (1x - 4 + √2)

2x³ - 6x² - 1x + (2-√2) = (2x³ - 1x) - (6x²

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