Question

Find the zeroes of the quadratic polynomial p(x)=6x2 - 3-7x​

Answer 1

Answer:

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Step-by-step explanation:

p(x)=6x²-7x-3

     =6x²-9x+2x-3

    =3x(2x-3)+1(2x-3)

    =(3x+1)(2x-3)

    x=-1/3,3/2

the zeroes of the formula are-1/3,3/2

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Answer 2

Answer:

$\huge\mathcal{AnsweR..}$

p(x)=> $6x{}^{2} - 7x -3=0$

By using spiliting middle term...

p(x)=>$6x{}^{2}-9x+2x-3=0$

= >3x(2x-3)+1(2x-3)

=> (2x-3) (3x+1)=0

=> 2x-3=0. , 3x+1 =0

=> $x = \frac{ - 3}{2} and x = \frac{ -1}{3}$

So

$\frac{ - 3}{2} and \frac{ -1}{3} are two zeros of following p(x)..$

Step-by-step explanation:

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