Question

How do you factor $2{{y}^{3}}+{{y}^{2}}-2y-1$?

Answer 1

Hint: For factoring the polynomial, which is given as $2{{y}^{3}}+{{y}^{2}}-2y-1$, we first have to guess its zero using the hit and trial method. For this, we will substitute $y=1$ in the given polynomial, which will give zero. By the factor theorem, we will obtain one factor as $\left( y-1 \right)$. Then we need to divide the given polynomial by the factor $\left( y-1 \right)$ using the long division method. The quotient will give the quadratic factor, which

Complete step by step solution:
Let us write the given polynomial as
\[p\left( y \right)=2{{y}^{3}}+{{y}^{2}}-2y-1\]
To start factorising, we need at least one zero of the given polynomial. For this, we use the hit and trial method and try substituting $y=1$ in the given polynomial to obtain
\[\begin{align}
  & \Rightarrow p\left( 1 \right)=2{{\left( 1 \right)}^{3}}+{{\left( 1 \right)}^{2}}-2\left( 1 \right)-1 \\
 & \Rightarrow p\left( 1 \right)=2+1-2-1 \\
 & \Rightarrow p\left( 1 \right)=3-3=0 \\
\end{align}\]
So $y=1$ is a zero of the given polynomial. By factor theorem, we can say that $\left( y-1 \right)$ is a factor of the given polynomial. So to obtain other factors, we divide the given polynomial by $\left( y-1 \right)$ as shown.
\[y-1\overset{2{{y}^{2}}+3y+1}{\overline{\left){\begin{align}
  & 2{{y}^{3}}+{{y}^{2}}-2y-1 \\
 & \underline{2{{y}^{3}}-2{{y}^{2}}} \\
 & 3{{y}^{2}}-2y-1 \\
 & \underline{3{{y}^{2}}-3y} \\
 & y-1 \\
 & \underline{y-1} \\
 & \underline{0} \\
\end{align}}\right.}}\]
From the above division, we can write the given polynomial as
$\Rightarrow p\left( y \right)=\left( y-1 \right)\left( 2{{y}^{2}}+3y+1 \right).......(i)$
Now, we take up the quadratic factor separately as
$\Rightarrow g\left( y \right)=2{{y}^{2}}+3y+1$
We factor it by using the middle term splitting, according to which we need to split the middle term into two terms such that their product is equal to the product of the first and the third terms, in this case equal to $\left( 2{{y}^{2}} \right)\left( 1 \right)=2{{y}^{2}}$. The only way to do this is to put $3y=2y+y$ above to get
$\Rightarrow g\left( y \right)=2{{y}^{2}}+2y+y+1$
Now, we take $2y$ common from the first two terms and $1$ common from the last two terms to get
\[\Rightarrow g\left( y \right)=2y\left( y+1 \right)+1\left( y+1 \right)\]
Now, taking \[\left( y+1 \right)\] common we get
$\Rightarrow g\left( y \right)=\left( y+1 \right)\left( 2y+1 \right)$
Now, we put this in (i) to get
$\Rightarrow p\left( y \right)=\left( y-1 \right)\left( y+1 \right)\left( 2y+1 \right)$
Hence, the given polynomial is factored.

Note:
We can also use the method of factor by grouping to factor the given polynomial very easily. For this, we need to divide the four terms given in the polynomial into two pairs and take factors common from each pair. For example, in the case of the given polynomial $2{{y}^{3}}+{{y}^{2}}-2y-1$, we will take ${{y}^{2}}$ common from the first two terms, and $-1$ common from the last two terms to get ${{y}^{2}}\left( 2y+1 \right)-1\left( 2y+1 \right)$. Finally, on taking $\left( 2y+1 \right)$ common and using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we will obtain the required factored form of the given polynomial.