Question

if the polynomial p(X)=x³-x²+3x+k is divided by (x-1) , the reminder obtained is 3 , then find the value of k​

Answer 1

Answer :-

Given :-

• p(x) = x³ - x² + 3x + k is divided by ( x - 1 )

• Remainder = 3

To Find :-

• The value of k

Solution :-

By using the remainder theorem :-

→ x - 1 = 0

→ x = 1

Substituting the value of x in polynomial and equating it to 3 :-

→ x³ - x² + 3x + k = 3

→ 1³ - 1² + 3 × 1 + k = 3

→ 1 - 1 + 3 + k = 3

→ 3 + k = 3

→ k = 3 - 3

→ k = 0

Value of k = 0

Answer 2

Given:

$\begin{gathered}\longmapsto\sf{Dividend=x^3-x^2+3x+k}\end{gathered}$

$\begin{gathered}\longmapsto\sf{Divisor=(x-1)}\end{gathered}$

$\begin{gathered}\longmapsto\sf{Remainder=3}\end{gathered}$

To find:

The value of k.

Solution:

$\begin{gathered}\longmapsto\sf{x-1 = 0}\end{gathered}$

$\begin{gathered}\longmapsto{\boxed{\pink{\frak{x= 1}}}}\end{gathered}$

$\bigstar$ Here, we are going to substitute the value of x = 1 in the given dividend.

$\bigstar$ By substituting, we will get the value of k, which is our required answer.

$\begin{gathered}\longmapsto\sf{f(x) = 3}\end{gathered}$

$\bigstar$ Why f(x) = 3? This is so because, the remainder given in the question is 3.

$\begin{gathered}\longmapsto\sf{f(x) = x^3-x^2+3x+k = 3}\end{gathered}$

$\begin{gathered}\longmapsto\sf{f(1) = 1^3-1^2+3(1)+k = 3}\end{gathered}$

$\begin{gathered}\longmapsto\sf{1-1+3+k = 3}\end{gathered}$

$\begin{gathered}\longmapsto\sf{ \cancel{1-1}+3+k = 3}\end{gathered}$

$\begin{gathered}\longmapsto\sf{3+k = 3}\end{gathered}$

$\begin{gathered}\longmapsto\sf{k = 3-3}\end{gathered}$

$\begin{gathered}\leadsto{\boxed{\red{\bf{k= 0}}}}\end{gathered}$

Hence, the value of k is 0