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Question: If \(\left( {x - 1} \right)\) is a factor of the polynomial \(3{x^3} - 2{x^2} + kx - 6\) , then find...

If (x1)\left( {x - 1} \right) is a factor of the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 , then find the value of kk ?

Explanation

Solution

If (x1)\left( {x - 1} \right) is a factor of the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6, then x=1x = 1 is one of its zeroes. it satisfies the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6. It means if we put x=1x = 1 in the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 then its value becomes equal to zero.

Complete step-by-step answer:
Here, the given polynomial is 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 and its one of the factors is (x1)\left( {x - 1} \right).
Since (x1)\left( {x - 1} \right) is a factor of the given polynomial x=1x = 1 satisfy the given polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6.
Now, put x=1x = 1in the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 and then equate the result with zero.
Putting x=1x = 1in the polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6. We get,
3(1)32(1)2+k(1)6=0 32+k6=0 k5=0 k=5  \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 kk is 55.

Note:
If aa, bb and cc are the zeroes of any cubic polynomial f(x)f\left( x \right), then (xa)\left( {x - a} \right) , (xb)\left( {x - b} \right) and (xc)\left( {x - c} \right) are the factors of the given polynomial f(x)f\left( x \right).
The given polynomial 3x32x2+kx63{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 kk 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 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 is given that is (x1)\left( {x - 1} \right). Now, divide the given polynomial 3x32x2+kx63{x^3} - 2{x^2} + kx - 6 by (x1)\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=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} where aa is the coefficient of x2{x^2}, bb is the coefficient of xx and cc is the constant term of the quadratic polynomial ax2+bx+ca{x^2} + bx + c.