Question

The three-degree polynomial \[f\left( x \right)\] has roots of the equation \[3, - 3\] and \[ - k\]. Given that the coefficient of \[{x^3}\] is 2 and \[f\left( x \right)\] has a remainder of 8 when divided by \[x + 1\]. The value of \[k\] is

Answer 1

Hint: First of all, form the cubic polynomial with the given roots. then use the formula if a polynomial \[f\left( x \right)\] has a remainder of \[r\] when divided by \[x - \alpha \] when \[f\left( \alpha \right) = r\] to find the required value of \[k\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that \[f\left( x \right)\] is a polynomial of degree three and its roots are \[3, - 3\] and \[ - k\].
Also given that \[f\left( x \right)\] has a remainder of 8 when divided by \[x + 1\].
We know that the equation of the cubic polynomial \[f\left( x \right)\] with roots \[\alpha ,\beta ,\gamma \] is given by \[f\left( x \right) = \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right) = 0\].
So, the given cubic polynomial \[f\left( x \right)\] with roots \[3, - 3\] and \[ - k\] is
\[
   \Rightarrow f\left( x \right) = \left( {x - 3} \right)\left( {x - \left( { - 3} \right)} \right)\left( {x - \left( { - k} \right)} \right) \\
   \Rightarrow f\left( x \right) = \left( {x - 3} \right)\left( {x + 3} \right)\left( {x + k} \right) \\
   \Rightarrow f\left( x \right) = \left( {{x^2} - 9} \right)\left( {x + k} \right) \\
\]
Also given that \[f\left( x \right)\] has a remainder of 8 when divided by \[x + 1\].
We know that if a polynomial \[f\left( x \right)\] has a remainder of \[r\] when divided by \[x - \alpha \] when \[f\left( \alpha \right) = r\]
Since \[f\left( x \right)\] has a remainder of 8 when divided by \[x + 1\], we have
\[
   \Rightarrow f\left( { - 1} \right) = 8 \\
   \Rightarrow \left( {{{\left( { - 1} \right)}^2} - 9} \right)\left( { - 1 + k} \right) = 8 \\
   \Rightarrow \left( {1 - 9} \right)\left( { - 1 + k} \right) = 8 \\
   \Rightarrow - 8\left( {k - 1} \right) = 8 \\
   \Rightarrow k - 1 = \dfrac{8}{{ - 8}} = - 1 \\
  \therefore k = - 1 + 1 = 0 \\
\]
Thus, the value of \[k\] is 0.

Note: A cubic polynomial is a polynomial of degree 3. A cubic polynomial is of the form \[a{x^3} + b{x^2} + cx + d\]. An equation involving a cubic polynomial is called as a cubic equation. A cubic equation is of the form \[a{x^3} + b{x^2} + cx + d = 0\].