Question

Using factor theorem, factorize each of the following polynomials: $2{{y}^{3}}-5{{y}^{2}}-19y+42$ .

Answer 1

Hint: To factorize $2{{y}^{3}}-5{{y}^{2}}-19y+42$ using factor theorem, we have to equate the given polynomial to f(y). Then, we have a trial and error method by substituting some values of y so that the above function will become 0. If \[f\left( -c \right)=0\] , then \[\left( y+c \right)\] is a factor of the polynomial f(y). If \[f\left( \dfrac{d}{c} \right)=\text{ }0\] then \[\left( cy-d \right)\] is a factor of the polynomial f(y). If \[f\left( -\dfrac{d}{c} \right)=\text{ }0\] , then \[\left( cy+d \right)\] is a factor of the polynomial f(y). . If \[f\left( c \right)=0\] , then \[\left( y-c \right)\] is a factor of the polynomial f(y). Then, we will divide the given polynomial by this factor. The quotient will be another polynomial. If we get the quotient as a quadratic polynomial, we have to further factorize this polynomial using splitting the middle terms or using the formula $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .

Complete step by step solution:
We have to factorize $2{{y}^{3}}-5{{y}^{2}}-19y+42$ using the factor theorem. We have to first equate the given polynomial to f(y).
$\Rightarrow f\left( y \right)=2{{y}^{3}}-5{{y}^{2}}-19y+42$
Now, we have to substitute some values of y so that the above function will become 0. This is the trial and error method.
Let us consider $y=0$ .
$\Rightarrow f\left( 0 \right)=2\times 0-5\times 0-19\times 0+42=42\ne 0$
Hence, this value of y cannot be taken.
Now, we have to consider $y=1$ .
$\begin{align}
  & \Rightarrow f\left( 1 \right)=2\times 1-5\times 1-19\times 1+42 \\
 & \Rightarrow f\left( 1 \right)=2-5-19+42=20\ne 0 \\
\end{align}$
Hence, this value of y cannot be taken.
Now, let us consider $y=2$
$\begin{align}
  & \Rightarrow f\left( 2 \right)=2{{\left( 2 \right)}^{3}}-5{{\left( 2 \right)}^{2}}-19\times 2+42 \\
 & \Rightarrow f\left( 2 \right)=16-20-38+42=0 \\
\end{align}$
We can see that $\left( y-2 \right)$ is a factor of $2{{y}^{3}}-5{{y}^{2}}-19y+42$ .
Now, we have to divide $2{{y}^{3}}-5{{y}^{2}}-19y+42$ by $\left( y-2 \right)$ .
\[y-2\overset{2{{y}^{2}}-y-21}{\overline{\left){\begin{align}
  & 2{{y}^{3}}-5{{y}^{2}}-19y+42 \\
 & 2{{y}^{3}}-4{{y}^{2}} \\
 & \left( - \right)\begin{matrix}
   {} & \left( + \right) \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & {} & -{{y}^{2}}-19y \\
\end{matrix} \\
 & \begin{matrix}
   {} & {} & -{{y}^{2}}+2y \\
\end{matrix} \\
 & \begin{matrix}
   {} & {} & \left( + \right)\begin{matrix}
   {} & \left( - \right) \\
\end{matrix} \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & {} & \begin{matrix}
   {} & {} & -21y+42 \\
\end{matrix} \\
\end{matrix} \\
 & \begin{matrix}
   {} & {} & \begin{matrix}
   {} & {} & -21y+42 \\
\end{matrix} \\
\end{matrix} \\
 & \begin{matrix}
   {} & {} & \begin{matrix}
   {} & {} & \left( + \right)\begin{matrix}
   {} & \left( - \right) \\
\end{matrix} \\
\end{matrix} \\
\end{matrix} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
 & \begin{matrix}
   {} & {} & \begin{matrix}
   {} & {} & \begin{matrix}
   {} & {} & 0 \\
\end{matrix} \\
\end{matrix} \\
\end{matrix} \\
\end{align}}\right.}}\]
Hence, we can write f(y) as
$f\left( y \right)=\left( y-2 \right)\left( 2{{y}^{2}}-y-21 \right)$
Now, we can factorize $2{{y}^{2}}-y-21$ by splitting the middle term. We can write –y as \[6y-7y\] as the sum is -1 and product is -21.
$\Rightarrow f\left( y \right)=\left( y-2 \right)\left( 2{{y}^{2}}+6y-7y-21 \right)$
Let us take the common factors outside.
$\Rightarrow f\left( y \right)=\left( y-2 \right)\left[ 2y\left( y+3 \right)-7\left( y+3 \right) \right]$
We can again take the common factors outside.
$\Rightarrow f\left( y \right)=\left( y-2 \right)\left( 2y-7 \right)\left( y+3 \right)$

Note: Students must know the long division method thoroughly to factorize the given polynomial. If \[f\left( -c \right)=0\] , then \[\left( y+c \right)\] is a factor of the polynomial f(y). If \[f\left( \dfrac{d}{c} \right)=\text{ }0\] then \[\left( cy-d \right)\] is a factor of the polynomial f(y). If \[f\left( -\dfrac{d}{c} \right)=\text{ }0\] , then \[\left( cy+d \right)\] is a factor of the polynomial f(y). . If \[f\left( c \right)=0\] , then \[\left( y-c \right)\] is a factor of the polynomial f(y). We can also use synthetic division instead of long division.