Remove the second term from the equation:
${x^4} + 4{x^3} + 2{x^2} - 4x - 2 = 0$
Answer
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Hint: In this particular question use the concept that in a fourth order polynomial the sum of the roots is the ratio of negative times the coefficient of ${x^3}$ to the coefficient of ${x^4}$, so in order to remove the second term (i.e. ${x^3}$ term) of the given polynomial, the sum of the roots of the polynomial must be zero so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given polynomial:
${x^4} + 4{x^3} + 2{x^2} - 4x - 2 = 0$
Now we have to remove the second term from the equation.
As we see that the highest power of x is 4, so it is a fourth order polynomial.
So this equation has at most 4 roots.
Now as we know that in a fourth order polynomial the sum of the roots is the ratio of negative times the coefficient of ${x^3}$ to the coefficient of ${x^4}$.
Let the roots of this equation be a, b, c and d.
So the sum of the roots is,
$a + b + c + d = \dfrac{{ - {\text{coefficient of }}{{\text{x}}^3}}}{{{\text{coefficient of }}{{\text{x}}^4}}}$
Therefore,
$a + b + c + d = \dfrac{{ - 4}}{1} = - 4$
Now in order to remove the second term (i.e. ${x^3}$ term) of the given polynomial, the sum of the roots of the polynomial must be zero.
Which can be achieved by increasing each of the roots of the given polynomial by 1.
So, the roots become, (a + 1), (b + 1), (c + 1) and (d + 1).
Therefore, transform (x – 1) in place of x in the given polynomial we have,
$ \Rightarrow {\left( {x - 1} \right)^4} + 4{\left( {x - 1} \right)^3} + 2{\left( {x - 1} \right)^2} - 4\left( {x - 1} \right) - 2 = 0$
Now simplify the above equation we have,
\[ \Rightarrow \left( {{x^2} + 1 - 2x} \right)\left( {{x^2} + 1 - 2x} \right) + 4\left( {{x^3} - 1 - 3{x^2} + 3x} \right) + 2\left( {{x^2} + 1 - 2x} \right) - 4\left( {x - 1} \right) - 2 = 0\]
\[ \Rightarrow \left( {{x^4} + {x^2} - 2{x^3} + {x^2} + 1 - 2x - 2{x^3} - 2x + 4{x^2}} \right) + \left( {4{x^3} - 4 - 12{x^2} + 12x} \right) + \left( {2{x^2} + 2 - 4x} \right) - \left( {4x - 4} \right) - 2 = 0\]
\[ \Rightarrow \left( {{x^4} - 4{x^3} + 6{x^2} - 4x + 1} \right) + \left( {4{x^3} - 4 - 12{x^2} + 12x} \right) + \left( {2{x^2} + 2 - 4x} \right) - \left( {4x - 4} \right) - 2 = 0\]
\[ \Rightarrow {x^4} - 4{x^2} + 1 = 0\]
So this is the required answer.
Note: Whenever we face such types of questions the key concept we have to remember is that as the sum of the roots of the polynomial is -4, so we have to increase the each roots of the given polynomial by 1 so that the sum of the roots become zero, so that second term is eliminated from the given polynomial, so we achieve thing by transform (x – 1) in place of x in the given polynomial and simplify as above we will get the required answer.
Complete step-by-step answer:
Given polynomial:
${x^4} + 4{x^3} + 2{x^2} - 4x - 2 = 0$
Now we have to remove the second term from the equation.
As we see that the highest power of x is 4, so it is a fourth order polynomial.
So this equation has at most 4 roots.
Now as we know that in a fourth order polynomial the sum of the roots is the ratio of negative times the coefficient of ${x^3}$ to the coefficient of ${x^4}$.
Let the roots of this equation be a, b, c and d.
So the sum of the roots is,
$a + b + c + d = \dfrac{{ - {\text{coefficient of }}{{\text{x}}^3}}}{{{\text{coefficient of }}{{\text{x}}^4}}}$
Therefore,
$a + b + c + d = \dfrac{{ - 4}}{1} = - 4$
Now in order to remove the second term (i.e. ${x^3}$ term) of the given polynomial, the sum of the roots of the polynomial must be zero.
Which can be achieved by increasing each of the roots of the given polynomial by 1.
So, the roots become, (a + 1), (b + 1), (c + 1) and (d + 1).
Therefore, transform (x – 1) in place of x in the given polynomial we have,
$ \Rightarrow {\left( {x - 1} \right)^4} + 4{\left( {x - 1} \right)^3} + 2{\left( {x - 1} \right)^2} - 4\left( {x - 1} \right) - 2 = 0$
Now simplify the above equation we have,
\[ \Rightarrow \left( {{x^2} + 1 - 2x} \right)\left( {{x^2} + 1 - 2x} \right) + 4\left( {{x^3} - 1 - 3{x^2} + 3x} \right) + 2\left( {{x^2} + 1 - 2x} \right) - 4\left( {x - 1} \right) - 2 = 0\]
\[ \Rightarrow \left( {{x^4} + {x^2} - 2{x^3} + {x^2} + 1 - 2x - 2{x^3} - 2x + 4{x^2}} \right) + \left( {4{x^3} - 4 - 12{x^2} + 12x} \right) + \left( {2{x^2} + 2 - 4x} \right) - \left( {4x - 4} \right) - 2 = 0\]
\[ \Rightarrow \left( {{x^4} - 4{x^3} + 6{x^2} - 4x + 1} \right) + \left( {4{x^3} - 4 - 12{x^2} + 12x} \right) + \left( {2{x^2} + 2 - 4x} \right) - \left( {4x - 4} \right) - 2 = 0\]
\[ \Rightarrow {x^4} - 4{x^2} + 1 = 0\]
So this is the required answer.
Note: Whenever we face such types of questions the key concept we have to remember is that as the sum of the roots of the polynomial is -4, so we have to increase the each roots of the given polynomial by 1 so that the sum of the roots become zero, so that second term is eliminated from the given polynomial, so we achieve thing by transform (x – 1) in place of x in the given polynomial and simplify as above we will get the required answer.
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