Find the zeroes of p(x)=9x³-x and hence verify the relationship betwe

1.	Find the zeroes of p(x)=9x³-x and hence verify the relationship between zeroes and coefficients.
Answer» ANSWER: $\sf{The \ zeroes \ of \ the \ polynomial \ are}$ $\sf{0, \ \dfrac{1}{3} \ and \ -\dfrac{1}{3}.}$ SOLUTION: $\sf{p(x)=9x^{3}-x}$ $\sf{\therefore{9x^{3}-x=0}}$ $\sf{\therefore{x(9x^{2}-1)=0}}$ $\sf{\therefore{x=0 \ or \ 9x^{2}-1=0}}$ $\sf{\therefore{x=0 \ or \ x^{2}=\dfrac{1}{9}}}$ $\sf{\therefore{x=0 \ or \ x=\dfrac{1}{3} \ or \ x=-\dfrac{1}{3}}}$ $\sf\purple{\tt{\therefore{The \ zeroes \ of \ the \ polynomial \ are}}}$ $\sf\purple{\tt{0, \ \dfrac{1}{3} \ and \ -\dfrac{1}{3}.}}$ _______________________________ $\sf\blue{Relation \ between \ coefficients \ and \ zeroes}$ $\sf{The \ given \ cubic \ polynomial \ is}$ $\sf{p(x)=9x^{3}-x}$ $\sf{In \ standard \ form}$ $\sf{\leadsto{p(x)=9x^{3}+0x^{2}-x+0}}$ $\sf{Here, \ a=9, \ b=0, \ c=-1 \ and \ d=0}$ $\sf{Let \ \alpha=0,}$ $\sf{\beta=\dfrac{1}{3},}$ $\sf{\gamma=-\dfrac{1}{3}}$ ______________________ $\sf{-\dfrac{b}{a}=-\dfrac{0}{9}}$ $\sf{\therefore{-\dfrac{b}{a}=0...(1)}}$ $\sf{\alpha+\beta+\gamma=\dfrac{1}{3}-\dfrac{1}{3}+0}$ $\sf{\therefore{\alpha+\beta+\gamma=0...(2)}}$ $\sf{from \ (1) \ and \ (2)}$ $\boxed{\sf{\alpha+\beta+\gamma=-\dfrac{b}{a}}}$ _______________________ $\sf{\dfrac{c}{a}=-\dfrac{1}{9}...(3)}$ $\sf{\alpha\beta+\beta\gamma+\alpha\gamma=(0)(\dfrac{1}{3})+(\dfrac{1}{3})(-\dfrac{1}{3})+(0)(-\dfrac{1}{3})}$ $\sf{\therefore{\alpha\beta+\beta\gamma+\alpha\gamma=-\dfrac{1}{9}...(4)}}$ $\sf{from \ (3) \ and \ (4)}$ $\boxed{\sf{\alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}}}$ _______________________ $\sf{-\dfrac{d}{a}=-\dfrac{0}{9}}$ $\sf{\therefore{-\dfrac{d}{a}=0...(5)}}$ $\sf{\alpha\beta\gamma=(0)(\dfrac{1}{3})(-\dfrac{1}{3})}$ $\sf{\therefore{\alpha\beta\gamma=0...(6)}}$ $\sf{from \ (5) \ and \ (6)}$ $\boxed{\sf{\alpha\beta\gamma=-\dfrac{d}{a}}}$