Question

Solve the equation \[2{{y}^{3}}\,+\,{{y}^{2}}\,-\,2y\,-1\]

Answer 1

Hint: First use hit and trial method, in which we will put \[-1,\text{ }+1,\text{ }-2,\]…..(factorization method)
Which number satisfies the equation that will be the factor of this equation. After that divide the equation with this factor you will get a quadratic equation. Then use the middle term split to get two factors.

Complete step-by-step answer:
\[2{{y}^{3}}\,+\,{{y}^{2}}\,-\,2y\,-1\]
Put 1
\[2{{(1)}^{3}}\,+\,{{(1)}^{2}}\,-\,2(1)\,-1\]
\[2\,+\,3\,-\,2\,-\,1\]
\[=\,0\]
So is the factor of this equation.
Now, \[\begin{align}
  & y\,=\,1 \\
 & y\,-1\,=\,0 \\
\end{align}\]

\[y-1\overset{2{{y}^{2}}+\,3y\,+1}{\overline{\left){\begin{align}
  & 2{{y}^{3}}\,+\,{{y}^{2}}\,-\,2y\,-1 \\
 & \dfrac{2{{y}^{3}}\,-\,2{{y}^{2}}}{\begin{align}
  & 3{{y}^{2}}\,-\,2y\,-\,1 \\
 & \dfrac{3{{y}^{2}}\,-\,3y}{\begin{align}
  & y\,-\,1 \\
 & \dfrac{y\,-\,1}{0} \\
\end{align}} \\
\end{align}} \\
 & \\
\end{align}}\right.}}\]
Now by dividing we get a quadratic equation:
\[2{{y}^{2}}\,+\,3y\,+\,1\]
\[\begin{align}
  & 2{{y}^{2}}+\,2y\,+\,y\,+\,1 \\
 & 2y\,(y+1)\,+1(y+1) \\
 & (2y+1)\,(y+1) \\
 & 2y\,=\,1\,\,\,\,\,\,\,\,y+1=0 \\
 & y\,=\,\dfrac{1}{2}\,\,\,\,\,\,\,\,\,y\,=\,-1 \\
 & \\
\end{align}\]
So there are three solutions of this question \[1,\,\dfrac{1}{2},\,-1\]

Note: If the power of question is three then there are three solutions to the question. In quadratic equations we can find a solution simply by the middle term split method but for cubic equations always use the hit and trial method and factorization method to find the answer.