Answer
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Hint:In the question, we are given that the $\left( {x - 1} \right)$ is a polynomial of the equation given. This means that if we put this value into the equation will give the value zero. So, put $x = 1$ into the equation and equate it to zero. Solve the above steps. Hence, we will find the answer.
Complete step by step solution:
In the above question, we are given that the $\left( {x - 1} \right)$ is a polynomial of $3{x^3} - 2{x^2} + kx - 6$ which means that if we equate the $\left( {x - 1} \right)$ to zero which gives
$
x - 1 = 0 \\
or \\
x = 1 \\
$
And then by putting this value to the polynomial then it will give us the value zero. Hence, putting the value into the polynomial
$3{\left( 1 \right)^3} - 2{\left( 1 \right)^2} + k\left( 1 \right) - 6 = 0$
Solving the above equation to find the value of k,
$
3 - 2 + k - 6 = 0 \\
\Rightarrow k = 5 \\
$
Hence, the solution for the question is $k = 5$
Note: Be careful while equating to zero. Because some students take $x + 1 = 0$ to be $x = 1$ which is totally wrong. These silly mistakes could lead to make the whole question wrong. While taking the values to the left-hand side the sign will change and vice-versa.
Complete step by step solution:
In the above question, we are given that the $\left( {x - 1} \right)$ is a polynomial of $3{x^3} - 2{x^2} + kx - 6$ which means that if we equate the $\left( {x - 1} \right)$ to zero which gives
$
x - 1 = 0 \\
or \\
x = 1 \\
$
And then by putting this value to the polynomial then it will give us the value zero. Hence, putting the value into the polynomial
$3{\left( 1 \right)^3} - 2{\left( 1 \right)^2} + k\left( 1 \right) - 6 = 0$
Solving the above equation to find the value of k,
$
3 - 2 + k - 6 = 0 \\
\Rightarrow k = 5 \\
$
Hence, the solution for the question is $k = 5$
Note: Be careful while equating to zero. Because some students take $x + 1 = 0$ to be $x = 1$ which is totally wrong. These silly mistakes could lead to make the whole question wrong. While taking the values to the left-hand side the sign will change and vice-versa.
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