
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
484.5k+ views
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 \[a{x^3} + b{x^2} + cx + d = 0\].
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 \[a{x^3} + b{x^2} + cx + d = 0\].
Recently Updated Pages
Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
