Find all the zeroes of 2x⁴-3xᶟ-3x²+6x-2, if you know that two of its zeroes are √2 and -√2. *

4 points

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2 thoughts on “Find all the zeroes of 2x⁴-3xᶟ-3x²+6x-2, if you know that two of its zeroes are √2 and -√2. *<br /><br />4 points<br /><br />Add F”

1. [tex] \sf{Solution}[/tex]

   GIVEN

   • √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2

   TO FIND

   • All the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2

   SOLUTION

   • √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2
   • So, (x+√2)(x-√2) i.e x² – 2 is the factor of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2.
   • By Dividing → (2x⁴ -3x³ – 3x² + 6x – 2) by (x² – 2) , I get Quotient = 2x² – 3x + 1

   By splitting middle term,

   • 2x² – 2x – x + 1 = 0
   • 2x ( x – 1 ) – 1 ( x – 1 ) = 0
   • (2x-1)(x-1) = 0
   • 2x-1 = 0 or x-1 = 0
   • x = 1/2 or 1

   Thus, all the zeroes of this polynomial, 2x⁴ – 3x³ – 3x² + 6x – 2 are 1, 1/2, √2 and -√2.

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2. GIVEN

   • √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2

   TO FIND

   • All the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2

   SOLUTION

   • √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2

   • So, (x+√2)(x-√2) i.e x² – 2 is the factor of this polynomial, 2x⁴ -3x³ – 3x² + 6x – 2.

   • By Dividing → (2x⁴ -3x³ – 3x² + 6x – 2) by (x² – 2) , I get Quotient = 2x² – 3x + 1

   By splitting middle term,

   • 2x² – 2x – x + 1 = 0
   • 2x ( x – 1 ) – 1 ( x – 1 ) = 0
   • (2x-1)(x-1) = 0
   • 2x-1 = 0 or x-1 = 0
   • x = 1/2 or 1

   Thus, all the zeroes of this polynomial, 2x⁴ – 3x³ – 3x² + 6x – 2 are 1, 1/2, √2 and -√2.

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