Question

Let \[f\left( x \right) = {x^{135}} + {x^{125}} - {x^{115}} + {x^5} + 1\] . If \[f\left( x \right)\] is divided by \[{x^3} - x\] then the remainder is some function of \[x\] say \[g\left( x \right)\] Find the value of \[g\left( {10} \right)\]

Answer 1

Hint: We will divide the given function by the cubic function so that the remainder will be the quadratic function. So we will write the given function in the form of quotient and remainder. Then we will substitute the value of the variable one by one. From there, we will find the value of constants and thus, we will get the required value.

Complete step-by-step answer:
The given function is
\[f\left( x \right)={{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1\] …………. $\left( 1 \right)$
When we will divide the given function by the cubic function , we will get the remainder as the quadratic equation.
Let the remainder obtained be:
\[g\left( x \right) = a{x^2} + bx + c\]
Now, we will write the given function in the form of quotient and remainder.
\[ \Rightarrow f\left( x \right) = \left( {{x^3} - x} \right)q\left( x \right) + a{x^2} + bx + c\] ……… \[\left( 2 \right)\]
Now, substituting the value of the variable \[x\] as 0 in equation \[\left( 1 \right)\], we get
\[ \Rightarrow f\left( 0 \right) = {0^{135}} + {0^{125}} - {0^{115}} + {0^5} + 1\]
\[ \Rightarrow f\left( 0 \right) = 1\] ……… $\left( 3 \right)$
Again, substituting the value of the variable as 0 in equation \[\left( 2 \right)\], we get
\[\Rightarrow f\left( 0 \right)=\left( {{0}^{3}}-0 \right)q\left( 0 \right)+a{{0}^{2}}+b0+c\]
On further simplification we get
\[\begin{array}{l} \Rightarrow f\left( 0 \right) = 0 + c\\ \Rightarrow f\left( 0 \right) = c\end{array}\]
Substituting the value of \[f\left( 0 \right)\] obtained in equation \[\left( 3 \right)\] in the above equation, we get
\[\Rightarrow c=1\]
Now, Substituting the value of the variable \[x\] as 1 in equation $\left( 1 \right)$, we get
\[\Rightarrow f\left( 1 \right)={{1}^{135}}+{{1}^{125}}-{{1}^{115}}+{{1}^{5}}+1\]
\[ \Rightarrow f\left( 1 \right) = 3\] ……… \[\left( 4 \right)\]
Again, substituting the value of the variable \[x\] as 1 in equation $\left( 2 \right)$, we get
\[\Rightarrow f\left( 0 \right)=\left( {{1}^{3}}-1 \right)q\left( 1 \right)+a{{\left( 1 \right)}^{2}}+b\cdot 1+c\]
On further simplification, we get
\[\begin{array}{l} \Rightarrow f\left( 1 \right) = 0 + a + b + c\\ \Rightarrow f\left( 1 \right) = a + b + c\end{array}\]
Substituting the value of \[f\left( 1 \right)\] obtained in equation \[\left( 4 \right)\] , we get
\[ \Rightarrow a + b + c = 3\]
Now, we will substitute the value of the constant \[c\] here
  \[\begin{array}{l} \Rightarrow a + b + 1 = 3\\ \Rightarrow a + b = 2\end{array}\]……… $\left( 5 \right)$
Now, we will substitute the value of the variable \[x\] as \[ - 1\] in equation \[\left( 1 \right)\].
\[ \Rightarrow f\left( 1 \right) = {\left( { - 1} \right)^{135}} + {\left( { - 1} \right)^{125}} - {\left( { - 1} \right)^{115}} + {\left( { - 1} \right)^5} + 1\]
\[\Rightarrow f\left( -1 \right)=-1\] ……… \[\left( 6 \right)\]
Again, substituting the value of the variable \[x\] as \[ - 1\] in equation $\left( 2 \right)$, we get
\[\Rightarrow f\left( 0 \right)=\left( {{\left( -1 \right)}^{3}}-\left( -1 \right) \right)q\left( -1 \right)~+a{{\left( -1 \right)}^{2}}+b.\left( -1 \right)+c\]
On further simplification, we get
\[\begin{array}{l} \Rightarrow f\left( { - 1} \right) = 0 + a - b + c\\ \Rightarrow f\left( { - 1} \right) = a - b + c\end{array}\]
Substituting the value of \[f\left( { - 1} \right)\] obtained in equation 6 , we get
\[ \Rightarrow a - b + c = - 1\]
Now, we will substitute the value of the constant \[c\] in the above equation, we get
  \[\begin{array}{l} \Rightarrow a - b + 1 = - 1\\ \Rightarrow a - b = - 2\end{array}\]……………… \[\left( 7 \right)\]
Adding equation 5 and equation 7, we get
\[ \Rightarrow a + b + a - b = 2 - 2\]
On further simplification, we get
\[ \Rightarrow 2a = 0 \Rightarrow a = 0\]
Now, we will substitute the value of \[a\] in equation 5, we get
\[\begin{array}{l} \Rightarrow 0 + b = 2\\ \Rightarrow b = 2\end{array}\]
Thus, the remainder after substituting the values becomes
\[\begin{array}{l} \Rightarrow g\left( x \right) = 0.{x^2} + 2.x + 1\\ \Rightarrow g\left( x \right) = 2x + 1\end{array}\]
Now, we will calculate the value of $g\left( 10 \right)$ .
$\Rightarrow g\left( 10 \right)=2\times 10+1$
On multiplying and then adding the numbers, we get
$\Rightarrow g\left( 10 \right)=21$
Hence, the value of \[g\left( {10} \right)\] is equal to 21.

Note: Since, we have used the cubic and quadratic equation here. We need to know some important properties of them. The number of roots of the polynomial is always equal to highest power in the polynomial. The number of roots of a quadratic polynomial is 2 because the highest power in a quadratic polynomial is 2 and similarly the number of roots of a cubic polynomial is 3 because the highest power in a cubic polynomial is 3.