Question

If two zeros of the polynomial $p\left( x \right)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ are $-\sqrt{2}$ and $\sqrt{2}$, then find out the other two zeroes.

Answer 1

Hint: Use the factor theorem of polynomials to determine at least one of the polynomial factors of the given polynomial. Then divide the given polynomial by the factor to find out other factors. Use the factor theorem again to determine the zeroes of the polynomial.

Complete step-by-step solution:
The factor theorem states that any polynomial $p\left( x \right)$ of degree $n\ge 1$ and any real number, $x-a$ is factor of $p\left( x \right)$ if and only if $p\left( a \right)=0$ that means $a$ is zero of $p\left( x \right)$ . Then we can express $p\left( x \right)$ as $q\left( x \right)\left( x-a \right)$ where $q\left( x \right)$ is a polynomial of degree $n-1$ \[\]
The given polynomial is \[\]
$p\left( x \right)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ \[\]
According to the data given in the question, $-\sqrt{2}$ and $\sqrt{2}$ are tow zeroes of $p\left( x \right)$. Now we use the factor theorem to conclude that $x-\left( -\sqrt{2} \right)=x+\sqrt{2}$ and $x-\sqrt{2}$ are factors of $p\left( x \right)$. So $\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)={{x}^{2}}-2$ is also a factor of $p\left( x \right)$.\[\]
Now we shall determine other factors of $p\left( x \right)$ by dividing $p\left( x \right)$ by ${{x}^{2}}-2$ using the log division method,
\[\begin{gathered}
  {x^2} - 2\underline {\mathop{\left){\vphantom{1\begin{gathered}
  2{x^4} - 3{x^3} - 3{x^2} + 6x - 2 \\
  2{x^4} - 4{x^2} \\
\end{gathered} }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered}
  2{x^4} - 3{x^3} - 3{x^2} + 6x - 2 \\
  2{x^4} - 4{x^2} \\
\end{gathered} }}}
\limits^{\displaystyle\,\,\, {2{x^2} - 3x + 1}}} \\
  {\text{ }}{\underline {\,
  \begin{gathered}
   - 3{x^3} + {x^2} + 6x - 2 \\
   - 3{x^3} + 6x \\
\end{gathered} \,}} \\
  {\text{ }}{\underline {\,
  \begin{gathered}
  {x^2} - 2 \ \\
  {x^2} - 2 \ \\
\end{gathered} \,}} \\
  {\text{ }}0 \\
\end{gathered} \]
So we have found another factor of $p\left( x \right)$ as $ 2{{x}^{2}}-3x+1=q\left( x \right) $ (say). Here we can express,\[\]
$p\left( x \right)=q\left( x \right)\left( {{x}^{2}}-2 \right)$ \[\]
 We are going further factorize $q\left( x \right)$ by splitting the middle term method.
\[\begin{align}
  & q\left( x \right)=2{{x}^{2}}-3x+1=2{{x}^{2}}-2x-x+1 \\
 & =2x\left( x-1 \right)-1\left( x-1 \right) \\
 & =\left( x-1 \right)\left( 2x-1 \right) \\
\end{align}\]
Putting the value of $q\left( x \right)$ in $p\left( x \right)$ we get
 $p\left( x \right)=\left( x-1 \right)\left( 2x-1 \right)\left( {{x}^{2}}-2 \right)$.
The zeros of $p\left( x \right)$ is determined by assigning $p\left( x \right)=0$.
\[\begin{align}
  & p\left( x \right)=\left( x-1 \right)\left( 2x-1 \right)\left( {{x}^{2}}-2 \right)=0 \\
 & \Rightarrow p\left( x \right)=1,\dfrac{1}{2},\sqrt{2,}-\sqrt{2} \\
\end{align}\]
So the other two zeros of the given polynomial are found to be 1 and $\dfrac{1}{2}$.

Note: You need to take care of the confusion between remainder and factor theorem. The factor theorem is used to find out zeros of polynomials. You need also to be careful during the division of polynomials. You can also use the synthetic division method to find out the roots.