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If one zero of quadric polynomial x square +(p+1)x-6 is 2,then what is the value of P?​

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If one zero of quadric polynomial x SQUARE +(p+1)x-6 is 2,then what is the value of P?

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Let the given polynomial be

\sf \: p(x) = {x}^{2} + (p + 1)x - 6

One of the zeros of the polynomial is 2

NoTE

Sum of Zeros : - x coefficient/x² coefficient

Product of Zeros : CONSTANT term /x² coefficient

Let the other zero be a

Here,

\tt \: 2a = - 6 \\ \\ \longrightarrow \: \boxed{\boxed{ \tt a = - 3}}

The zeros of the polynomial are 2 and - 3

Also,

\tt \: \longrightarrow 2 + ( - 3) = - \dfrac{(p + 1)}{1} \\ \\ \longrightarrow \: \tt \: -1 = - (p + 1) \\ \\ \large{\longrightarrow\: \boxed{ \boxed{ \tt \: p = 0}}}


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