Question

To what expression must \[99{x^3} - 33{x^2} - 13x - 41\] be added to make the sum zero?

Answer 1

Hint: In the above given problem, we are given an expression as \[99{x^3} - 33{x^2} - 13x - 41\] . We have to find another expression for the given expression, such that adding \[99{x^3} - 33{x^2} - 13x - 41\] to that new expression gives the sum equal to zero. Since the given expression \[99{x^3} - 33{x^2} - 13x - 41\] is a function of the variable \[x\] , therefore suppose an another function of the same variable \[x\] , such that their sum if zero. Then substituting the value of the first function will give us the value of the required function.

Complete step by step answer:
Given expression is \[99{x^3} - 33{x^2} - 13x - 41\]. We have to find another expression such that the sum of both the expressions becomes zero. Since the given expression \[99{x^3} - 33{x^2} - 13x - 41\] is a function of the variable \[x\]. Therefore let,
\[ \Rightarrow f\left( x \right) = 99{x^3} - 33{x^2} - 13x - 41\]
Now consider another function of \[x\] as \[g\left( x \right)\]. According to the question,
\[ \Rightarrow f\left( x \right) + g\left( x \right) = 0\]

That gives us,
\[ \Rightarrow g\left( x \right) = - f\left( x \right)\]
Putting the value of the function \[f\left( x \right)\] we can write the above equation as
\[ \Rightarrow g\left( x \right) = - \left( {99{x^3} - 33{x^2} - 13x - 41} \right)\]
Removing the bracket, we get
\[ \therefore g\left( x \right) = - 99{x^3} + 33{x^2} + 13x + 41\]
That is the required function of \[x\]. Here, the expression of the function \[g\left( x \right)\] is \[ - 99{x^3} + 33{x^2} + 13x + 41\].

Therefore, the expression \[ - 99{x^3} + 33{x^2} + 13x + 41\] must be added to the expression \[99{x^3} - 33{x^2} - 13x - 41\] to make the sum zero.

Note: We can also verify our solution by substituting the value of the functions \[f\left( x \right)\] and \[g\left( x \right)\] in the equation \[f\left( x \right) + g\left( x \right) = 0\]. After substitution of the two functions \[f\left( x \right)\] and \[g\left( x \right)\] , we can write the above equation as,
\[ \Rightarrow f\left( x \right) + g\left( x \right) = \left( {99{x^3} - 33{x^2} - 13x - 41} \right) + \left( { - 99{x^3} + 33{x^2} + 13x + 41} \right)\]
Removing the brackets from both the expressions in the RHS, we get
\[ \Rightarrow f\left( x \right) + g\left( x \right) = 99{x^3} - 33{x^2} - 13x - 41 - 99{x^3} + 33{x^2} + 13x + 41\]
Now taking the like terms together, we can write
\[ \Rightarrow f\left( x \right) + g\left( x \right) = \left( {99{x^3} - 99{x^3}} \right) + \left( { - 33{x^2} + 33{x^2}} \right) + \left( { - 13x + 13x} \right) + \left( { - 41 + 41} \right)\]
Since all the terms are additive inverse of each other, therefore their individual sum inside the bracket will be equal to zero.
\[ \Rightarrow f\left( x \right) + g\left( x \right) = \left( 0 \right) + \left( 0 \right) + \left( 0 \right) + \left( 0 \right)\]
Therefore, we have
\[ \therefore f\left( x \right) + g\left( x \right) = 0\]
Hence, our solution is verified.