
Obtain all the values of \[f\left( x \right)=2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6\], if two of its zeros are \[\sqrt{\dfrac{3}{2}}\]and \[-\sqrt{\dfrac{3}{2}}\].
Answer
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Hint: Given that the function is \[f\left( x \right)=2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6\]. Two of its roots are\[\sqrt{\dfrac{3}{2}}\]and \[-\sqrt{\dfrac{3}{2}}\] . We have to solve the function with the Division algorithm method which is the same as normal division. It is step by step procedure to eliminate all the terms and write them in the form if
Dividend = Divisor \[\times \] Quotient + Remainder.
Complete step by step answer:
The given function is \[f\left( x \right)=2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6\] .
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.
\[({{x}^{2}}-\dfrac{3}{2})\overset{2{{x}^{2}}-2x-4}{\overline{\left){2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6}\right.}}\]
\[2{{x}^{4}}\] \[-3{{x}^{2}}\]
… … … … … … … … … … … … … … … … … … … … …..
\[-2{{x}^{3}}-4{{x}^{2}}+3x\]
\[-2{{x}^{3}}\] \[+3x\]
… … … … … … … … … … ………………………………..
\[-4{{x}^{2}}+6\]
\[-4{{x}^{2}}+6\]
…………………………………………………………….
0
We have got \[2{{x}^{2}}-2x-4\] as quotient. Finding the roots of\[2{{x}^{2}}-2x-4\].
\[2{{x}^{2}}-2x-4\]= \[{{x}^{2}}-x-2\]
\[\begin{align}
& {{x}^{2}}-2x+x-2 \\
& x\left( x-2 \right)+1\left( x-2 \right) \\
& \left( x=2 \right),\left( x=-1 \right) \\
\end{align}\]
Therefore zeros of the above function are \[\left( x=2 \right),\left( x=-1 \right)\].
Note: A proven statement that we use to prove another statement. On the other hand, an algorithm refers to a series of well-defined steps that gives a procedure for solving a type of problem.
Dividend = Divisor \[\times \] Quotient + Remainder.
Complete step by step answer:
The given function is \[f\left( x \right)=2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6\] .
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.
\[({{x}^{2}}-\dfrac{3}{2})\overset{2{{x}^{2}}-2x-4}{\overline{\left){2{{x}^{4}}-2{{x}^{3}}-7{{x}^{2}}+3x+6}\right.}}\]
\[2{{x}^{4}}\] \[-3{{x}^{2}}\]
… … … … … … … … … … … … … … … … … … … … …..
\[-2{{x}^{3}}-4{{x}^{2}}+3x\]
\[-2{{x}^{3}}\] \[+3x\]
… … … … … … … … … … ………………………………..
\[-4{{x}^{2}}+6\]
\[-4{{x}^{2}}+6\]
…………………………………………………………….
0
We have got \[2{{x}^{2}}-2x-4\] as quotient. Finding the roots of\[2{{x}^{2}}-2x-4\].
\[2{{x}^{2}}-2x-4\]= \[{{x}^{2}}-x-2\]
\[\begin{align}
& {{x}^{2}}-2x+x-2 \\
& x\left( x-2 \right)+1\left( x-2 \right) \\
& \left( x=2 \right),\left( x=-1 \right) \\
\end{align}\]
Therefore zeros of the above function are \[\left( x=2 \right),\left( x=-1 \right)\].
Note: A proven statement that we use to prove another statement. On the other hand, an algorithm refers to a series of well-defined steps that gives a procedure for solving a type of problem.
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