Find the value of 'p' for which -4 is a zero of x^{2} -2x-(7p+3)

1.	Find the value of 'p' for which -4 is a zero of x^{2} -2x-(7p+3)
Answer» $\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Question}}}}}{\bigstar}$ □☆□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆ $Find \ the \ value \ of \ 'p ' \ for \ which \ -4 \ is \ a \\ zero \ of \ x^{2} -2x-(7p+3)$ □☆□☆□☆□☆□☆□☆□☆□☆□☆□☆□☆ ────────────────────── $\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$ $\pink{\tt{Let \ other \ zero \beta}}$ $\boxed{\red{\bold{Sum \ of \ zeroes:}}}$ $\red{\implies}-4 + \beta = \displaystyle \frac{-b}{a}$ $\red{\implies}-4 + \beta = \displaystyle \frac{-(-2)}{1}$ $\red{\implies}-4 + \beta = 2$ $\red{\implies}\beta = 2+4 = 6$ $\boxed{\red{\bold{Product\ of \ zeroes:}}}$ $\green{\implies}-4 × \beta = \displaystyle \frac{c}{a}$ $\green{\implies}-4 × 6 = \displaystyle\frac{-7p-3}{1}$ $\green{\implies}-24 = -7p -3$ $\green{\implies}7p = 24-3$ $\green{\implies}7p = 21$ $\green{\boxed{\implies{\boxed{p =3}}}}$ ──────────────────────

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