Question

Find all the zeros of \[p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\] , if you know that two of its zeros are \[\sqrt{2}\]and \[-\sqrt{2}\].

Answer 1

Hint: Given that the function is \[p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\] and given zeros are \[\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=\left( {{x}^{2}}-2 \right)\] . We will perform the long division method for polynomials using the basic methodology, Dividend = Divisor * Quotient + Remainder, where dividend is the given function p(x), divisor is the product of the given zeros of the function p(x), quotient is the answer we get after division and remainder is any remaining term or number after division.

Complete step by step answer:
The given function is\[p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\] . The roots of the given function is \[\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=\left( {{x}^{2}}-2 \right)\]
First arrange the term of dividend and the divisor in the decreasing order of their degrees.
To obtain the first term of the quotient divide the highest degree term of the dividend by the highest degree term of the divisor.
To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor.
Continue this process till the degree of remainder is less than the degree of divisor.
\[\begin{align}
  & {{x}^{2}}-2\overset{\,\,2{{x}^{2}}-3x+1}{\overline{\left){2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2}\right.}} \\
 & \,\,\,\,\,\,\,\,\,\,-\underline{\left( 2{{x}^{4}} \right)\,\,\,\,\,\,\,+\left( -4{{x}^{2}} \right)} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3{{x}^{3}}+\,\,\,\,{{x}^{2}}+6x \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( -\,3{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6x \right)} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\underline{\left( {{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2 \right)} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
Finding the roots of \[2{{x}^{2}}-3x+1\] as,
\[\begin{align}
  & 2{{x}^{2}}-3x+1=2{{x}^{2}}-2x-x+1 \\
 & =2x\left( x-1 \right)-1\left( x-1 \right) \\
 & =\left( 2x-1 \right)\left( x-1 \right)
\end{align}\]
Put the above expression equal to 0 as,
\[\begin{align}
  & \left( 2x-1 \right)=0 \\
 & 2x=1 \\
 & x=\dfrac{1}{2}
\end{align}\]
  \[\begin{align}
  & \left( x-1 \right)=0 \\
 & x=1
\end{align}\]

Therefore the remaining zeros of the above function \[p(x)=2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\] are \[\dfrac{1}{2}\] and 1.

Note: The possible error that you may encounter could be that the long division was not done properly. Be careful when calculating the roots and keep in mind the sign of the expression being operated on.