Question

two zeroes of the polynomial p(x) x^3-4x^2+x+6 are 2 and -1 find the third zero​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{p(x) = {x}^{3} - {4x}^{2} + x + 6} \\ &\sf{ \blue{\alpha = 2}}\\ &\sf{ \green{ \beta = - \: 1}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{ \red{third \: zero \: of \: polynomial}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  Let \: the \:   \begin{cases} &\sf{ \purple{third \: zero \: be \: \gamma }}  \end{cases}\end{gathered}\end{gathered}$

Now,

We know that,

$\tt \:  \longrightarrow \: \boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

OR

$\tt \:  ⟼ \boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\tt\implies \: \alpha + \beta + \gamma = - \dfrac{( - 4)}{1}$

$\tt\implies \:2 \: + \: ( - 1) + \gamma = 4$

$\tt\implies \:1 + \gamma = 4$

$\tt\implies \: \gamma = 3$

$\tt\implies \: \boxed{ \mathcal{ \purple{third \: zero \: of \: polynomial \: is \: 3}}}$