The transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is
(A) ${x^4} - 24{x^2} + 65x - 55 = 0$
(B) ${x^4} + 24{x^2} + 65x + 55 = 0$
(C) ${x^4} - 24{x^2} - 65x + 55 = 0$
(D) ${x^4} + 24{x^2} + 65x - 55 = 0$
Answer
649.5k+ views
Hint: At first use the substitution \[\left( {x + h} \right)\] in the place of x in the given equation. Now note that, in order to eliminate the second term from this transformed equation, the coefficient of \[{x^3}\] must be zero. Therefore find h and rewrite the transformed equation substituting the value of h.
Complete step by step answer:
According to the question we are to eliminate the second term from the equation ${x^4} + 8{x^3} + x - 5 = 0$..... (1)
Let us at first use the substitution \[\left( {x + h} \right)\] in the place of x in equation (1).
Therefore we have,
${\left( {x + h} \right)^4} + 8{\left( {x + h} \right)^3} + \left( {x + h} \right) - 5 = 0$
On expanding we get,
$ \Rightarrow {x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4} + 8\left[ {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right] + x + h - 5 = 0$
On taking terms common we get,
$ \Rightarrow {x^4} + \left( {4h + 8} \right){x^3} + \left( {6{h^2} + 24h} \right){x^2} + \left( {4{h^3} + 24{h^2} + 1} \right)x + \left( {{h^4} + 8{h^3} + h - 5} \right) = 0$ ..... (2)
Now, in order to eliminate the second term from this transformed equation of${x^4} + 8{x^3} + x - 5 = 0$, we have to make the coefficient of \[{x^3}\] zero.
$ \Rightarrow 4h + 8 = 0$
On cross multiplication and simplification we get,
$ \Rightarrow h = \dfrac{{ - 8}}{4} = - 2$
Now, putting h=−2 in (2) we get,
$ \Rightarrow {x^4} + \left( { - 8 + 8} \right){x^3} + \left( {24 - 48} \right){x^2} + \left( { - 32 + 96 + 1} \right)x + \left( {16 - 64 - 2 - 5} \right) = 0$
On simplification we get,
$ \Rightarrow {x^4} - 24{x^2} + 65x - 55 = 0$
Therefore, the transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is ${x^4} - 24{x^2} + 65x - 55 = 0$
Hence, option (A) is correct.
Note: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. An example of a polynomial of a single indeterminate x is \[{x^2}\; - {\text{ }}4x{\text{ }} + {\text{ }}7\]. Here, we have to eliminate second term so we take coefficient of \[{x^3}\] to be zero, in other questions if it is asked to eliminate any other term we first substitute \[\left( {x + h} \right)\] in place of x, and then take the coefficient of the term which is asked to eliminate, as 0, and then we solve further.
Complete step by step answer:
According to the question we are to eliminate the second term from the equation ${x^4} + 8{x^3} + x - 5 = 0$..... (1)
Let us at first use the substitution \[\left( {x + h} \right)\] in the place of x in equation (1).
Therefore we have,
${\left( {x + h} \right)^4} + 8{\left( {x + h} \right)^3} + \left( {x + h} \right) - 5 = 0$
On expanding we get,
$ \Rightarrow {x^4} + 4{x^3}h + 6{x^2}{h^2} + 4x{h^3} + {h^4} + 8\left[ {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right] + x + h - 5 = 0$
On taking terms common we get,
$ \Rightarrow {x^4} + \left( {4h + 8} \right){x^3} + \left( {6{h^2} + 24h} \right){x^2} + \left( {4{h^3} + 24{h^2} + 1} \right)x + \left( {{h^4} + 8{h^3} + h - 5} \right) = 0$ ..... (2)
Now, in order to eliminate the second term from this transformed equation of${x^4} + 8{x^3} + x - 5 = 0$, we have to make the coefficient of \[{x^3}\] zero.
$ \Rightarrow 4h + 8 = 0$
On cross multiplication and simplification we get,
$ \Rightarrow h = \dfrac{{ - 8}}{4} = - 2$
Now, putting h=−2 in (2) we get,
$ \Rightarrow {x^4} + \left( { - 8 + 8} \right){x^3} + \left( {24 - 48} \right){x^2} + \left( { - 32 + 96 + 1} \right)x + \left( {16 - 64 - 2 - 5} \right) = 0$
On simplification we get,
$ \Rightarrow {x^4} - 24{x^2} + 65x - 55 = 0$
Therefore, the transformed equation of ${x^4} + 8{x^3} + x - 5 = 0$ by eliminating second term is ${x^4} - 24{x^2} + 65x - 55 = 0$
Hence, option (A) is correct.
Note: In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. An example of a polynomial of a single indeterminate x is \[{x^2}\; - {\text{ }}4x{\text{ }} + {\text{ }}7\]. Here, we have to eliminate second term so we take coefficient of \[{x^3}\] to be zero, in other questions if it is asked to eliminate any other term we first substitute \[\left( {x + h} \right)\] in place of x, and then take the coefficient of the term which is asked to eliminate, as 0, and then we solve further.
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