Question

Obtain all other zeros of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeros are √(5/3) and -√(5/3) .

Answer 1

Dear Student.

To find: zeros of 3x⁴ + 6x³ – 2x² – 10x – 5,

Given: two of its zeros are √(5/3) and -√(5/3) .

See the attachment.

Answer 2

Solution :

Given p(x) = 3x⁴+6x³-2x²-10x-5--(1)

The degree of p(x) is 4 , so it has

atmost 4 zeroes.

The two zeroes of given p(x) are

√(5/3) and -√(5/3)

The equation of the polynomial

whose zeroes are √(5/3) and -√(5/3 )

= ( x+√5/3 )( x - √5/3 )

= x² - 5/3 --- (2)

Divide (1) with ( 2 ) , we get

x²-5/3)3x⁴+6x³-2x²-10x-5(3x²+6x+3
********3x⁴+ 0 - 5x²
___________________
*************6x³+3x²-10x-5
*************6x³ + 0 - 10x
___________________
******************3x² + 0 - 5
******************3x² + 0 - 5
____________________
******************* ( 0 )

p(x) = ( x²-5/3)( 3x²+6x+3 )

= 3(x²-5/3)( x²+2x+1 )

= 3[ x² - {√(5/3)}² ] ( x + 1 )²

= 3(x+√5/3)(x-√5/3)(x+1)(x+1)

Therefore ,

The zeroes of the given polynomial

are -√(5/3), √(5/3) , -1, -1

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