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

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 $a{x^3} + b{x^2} + cx + d = 0$.