Math, asked by Ashwinchandrakar, 10 months ago

Obtain all other zeroes of polynomial 2x4+3x³-5x²-9x-3, if 2 of its zeros are √3 and -√3​


gsvk33: can u provide question once more
Ashwinchandrakar: Obtain all other zeroes of polynomial 2x4+3x³-5x²-9x-3, if 2 of its zeros are √3 and -√3
gsvk33: see ur answer bro

Answers

Answered by gsvk33
38

Answer:

the other zeroes are -1/2 and -1

Step-by-step explanation:

given polynomial : 2x⁴+3x³-5x²-9x-3

2 of its zeroes are +√3 and -√3

let the zeroes be (x+√3)(x-√3)

hence zeroes = x²-(√3)²

=x²-3

now dividing polynomial by x²-3..

x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1

       2x⁴         -6x²

       ⁽⁻⁾             ⁽⁺⁾

     ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

             3x³+ x²-9x

             3x        -9x

           ⁽⁻⁾             ⁽⁺⁾

           ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

                   x²-3

                   x²-3

                  ⁽⁻⁾   ⁽⁺⁾

               ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻              

                     0  0

now factorising, 2x²+3x+1

2x²+2x+1x+1

2x(x+1)+1(x+1)

(2x+1)(x+1)=0

2x+1=0                x+1=0

2x= -1                  x= -1

x = -1/2

therefore other zeroes are -1/2 and -1

hope its correct

mark me as brainliest

Answered by khushipachua
3

Answer:

the other zeroes are -1/2 and -1

given polynomial : 2x⁴+3x³-5x²-9x-3

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x             3x        -9x

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x             3x        -9x           ⁽⁻⁾             ⁽⁺⁾

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x             3x        -9x           ⁽⁻⁾             ⁽⁺⁾                   x²-3

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x             3x        -9x           ⁽⁻⁾             ⁽⁺⁾                   x²-3                   x²-3

given polynomial : 2x⁴+3x³-5x²-9x-32 of its zeroes are +√3 and -√3let the zeroes be (x+√3)(x-√3)hence zeroes = x²-(√3)²=x²-3now dividing polynomial by x²-3..x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1       2x⁴         -6x²       ⁽⁻⁾             ⁽⁺⁾             3x³+ x²-9x             3x        -9x           ⁽⁻⁾             ⁽⁺⁾                   x²-3                   x²-3                  ⁽⁻⁾   (+)         

                              0  0

                              0  0now factorising, 2x²+3x+1

                              0  0now factorising, 2x²+3x+12x²+2x+1x+1

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=0

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=02x+1=0                x+1=0

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=02x+1=0                x+1=02x= -1                  x= -1

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=02x+1=0                x+1=02x= -1                  x= -1x = -1/2

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=02x+1=0                x+1=02x= -1                  x= -1x = -1/2therefore other zeroes are -1/2 and -1

                              0  0now factorising, 2x²+3x+12x²+2x+1x+12x(x+1)+1(x+1)(2x+1)(x+1)=02x+1=0                x+1=02x= -1                  x= -1x = -1/2therefore other zeroes are -1/2 and -1hope its correct

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