Question

Find all the zeros of 2x⁴-3x³-5x²+9x-3,it being given that two of zeros are√3 and-√3



1 & 1/2

2

3/2,&6/2

√2 &-√2


​

Answer 1

p(x) : 2x⁴+3x³-5x²-9x-3

zeroes are √3 and -√3

So factors will be (x+√3)(x-√3)

hence zeroes = x²-(√3)²

=x²-3

g(x) = x²-3

x²-3) 2x⁴+3x³-5x²-9x-3 ( 2x²+3x+1

2x⁴ -6x²

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

3x³+ x²-9x

3x -9x

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

x²-3

x²-3

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

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

Answer 2

Answer:

Step-by-step explanation:

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 iy helps

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