
If $\left( {x - 1} \right)$ is a factor of the polynomial $3{x^3} - 2{x^2} + kx - 6$ , then find the value of $k$ ?
Answer
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Hint: If $\left( {x - 1} \right)$ is a factor of the polynomial $3{x^3} - 2{x^2} + kx - 6$, then $x = 1$ is one of its zeroes. it satisfies the polynomial $3{x^3} - 2{x^2} + kx - 6$. It means if we put $x = 1$ in the polynomial $3{x^3} - 2{x^2} + kx - 6$ then its value becomes equal to zero.
Complete step-by-step answer:
Here, the given polynomial is $3{x^3} - 2{x^2} + kx - 6$ and its one of the factors is $\left( {x - 1} \right)$.
Since $\left( {x - 1} \right)$ is a factor of the given polynomial $x = 1$ satisfy the given polynomial $3{x^3} - 2{x^2} + kx - 6$.
Now, put $x = 1$in the polynomial $3{x^3} - 2{x^2} + kx - 6$ and then equate the result with zero.
Putting $x = 1$in the polynomial $3{x^3} - 2{x^2} + kx - 6$. We get,
$
\Rightarrow 3{\left( 1 \right)^3} - 2{\left( 1 \right)^2} + k\left( 1 \right) - 6 = 0 \\
\Rightarrow 3 - 2 + k - 6 = 0 \\
\Rightarrow k - 5 = 0 \\
\therefore k = 5 \\
$
Thus, the required value of $k$ is $5$.
Note:
If \[a\], $b$ and $c$ are the zeroes of any cubic polynomial $f\left( x \right)$, then $\left( {x - a} \right)$ , $\left( {x - b} \right)$ and $\left( {x - c} \right)$ are the factors of the given polynomial $f\left( x \right)$.
The given polynomial $3{x^3} - 2{x^2} + kx - 6$ is a cubic polynomial and its one of the zeroes is given to us then with this data we can find the value of $k$ to get the required polynomial. Since this is a cubic it has three zero but only one zero is known so, its other two zeroes can be calculated as:
One of the factors of the given polynomial $3{x^3} - 2{x^2} + kx - 6$ is given that is $\left( {x - 1} \right)$. Now, divide the given polynomial $3{x^3} - 2{x^2} + kx - 6$ by $\left( {x - 1} \right)$ and it will give a quadratic polynomial then we have to calculate the zeros of a quadratic equation which can be calculated by using a well-known formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term of the quadratic polynomial $a{x^2} + bx + c$.
Complete step-by-step answer:
Here, the given polynomial is $3{x^3} - 2{x^2} + kx - 6$ and its one of the factors is $\left( {x - 1} \right)$.
Since $\left( {x - 1} \right)$ is a factor of the given polynomial $x = 1$ satisfy the given polynomial $3{x^3} - 2{x^2} + kx - 6$.
Now, put $x = 1$in the polynomial $3{x^3} - 2{x^2} + kx - 6$ and then equate the result with zero.
Putting $x = 1$in the polynomial $3{x^3} - 2{x^2} + kx - 6$. We get,
$
\Rightarrow 3{\left( 1 \right)^3} - 2{\left( 1 \right)^2} + k\left( 1 \right) - 6 = 0 \\
\Rightarrow 3 - 2 + k - 6 = 0 \\
\Rightarrow k - 5 = 0 \\
\therefore k = 5 \\
$
Thus, the required value of $k$ is $5$.
Note:
If \[a\], $b$ and $c$ are the zeroes of any cubic polynomial $f\left( x \right)$, then $\left( {x - a} \right)$ , $\left( {x - b} \right)$ and $\left( {x - c} \right)$ are the factors of the given polynomial $f\left( x \right)$.
The given polynomial $3{x^3} - 2{x^2} + kx - 6$ is a cubic polynomial and its one of the zeroes is given to us then with this data we can find the value of $k$ to get the required polynomial. Since this is a cubic it has three zero but only one zero is known so, its other two zeroes can be calculated as:
One of the factors of the given polynomial $3{x^3} - 2{x^2} + kx - 6$ is given that is $\left( {x - 1} \right)$. Now, divide the given polynomial $3{x^3} - 2{x^2} + kx - 6$ by $\left( {x - 1} \right)$ and it will give a quadratic polynomial then we have to calculate the zeros of a quadratic equation which can be calculated by using a well-known formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term of the quadratic polynomial $a{x^2} + bx + c$.
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