Question

If one zero of the polynomial $P\left( x \right) = {x^3} - 6{x^2} + 11x - 6$ is 3, find the other two zeroes.
$
  (a){\text{ 0 and 2}} \\
  {\text{(b) 2 and - 2}} \\
  {\text{(c) 1 and 2}} \\
  {\text{(d) 2 and - 3}} \\
 $

Answer 1

Hint – Since the given polynomial is cubic thus it must have 3 roots, as 3 is a root thus $x - 3 = 0$ must satisfy the so write the given polynomial in the form that $\dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{x - 3}} = a{x^2} + bx + c$. Simplify to find the value of a, b and c and thus eventually quadratic equation $a{x^2} + bx + c$ will be obtained. This quadratic when solved will give the other two roots.

Complete step-by-step answer:
Given polynomial
$P\left( x \right) = {x^3} - 6{x^2} + 11x - 6$
As we know that the zeros of the polynomial are nothing but the roots of the polynomial.
Now it is given that one of the roots of the polynomial is 3.
We have to find out the other two roots, as the given polynomial is a cubic polynomial.
Now if 3 is the root of the polynomial then (x – 3) = 0 is one of the factor of the polynomial so divide the cubic polynomial by this factor we have,
Let the other factor of the polynomial is $a{x^2} + bx + c = 0$
$ \Rightarrow \dfrac{{{x^3} - 6{x^2} + 11x - 6}}{{x - 3}} = a{x^2} + bx + c$
$ \Rightarrow {x^3} - 6{x^2} + 11x - 6 = \left( {x - 3} \right)\left( {a{x^2} + bx + c} \right)$
Now simplify we have,
$ \Rightarrow {x^3} - 6{x^2} + 11x - 6 = a{x^3} + b{x^2} + cx - 3a{x^2} - 3bx - 3c$
$ \Rightarrow {x^3} - 6{x^2} + 11x - 6 = a{x^3} + {x^2}\left( {b - 3a} \right) + x\left( {c - 3b} \right) - 3c$
Now compare the terms we have,
$a = 1..............\left( 1 \right)$
$\left( {b - 3a} \right) = - 6..............\left( 2 \right)$
$\left( {c - 3b} \right) = 11..................\left( 3 \right)$
$3c = 6...................\left( 4 \right)$
Now from equation (4) we have,
$ \Rightarrow c = \dfrac{6}{3} = 2$
Now substitute this value in equation (3) we have,
$\left( {2 - 3b} \right) = 11$
$ \Rightarrow 3b = 2 - 11 = - 9$
$ \Rightarrow b = \dfrac{{ - 9}}{3} = - 3$
So the other factor of the quadratic equation is
$a{x^2} + bx + c = {x^2} - 3x + 2 = 0$
Now factorize this equation we have,
$ \Rightarrow {x^2} - x - 2x + 2 = 0$
$ \Rightarrow x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0$
$ \Rightarrow \left( {x - 2} \right)\left( {x - 1} \right) = 0$
$ \Rightarrow x = 1,2$
So the other two factors are 1 and 2.
So this is the required answer.
Hence option (C) is the correct answer.

Note – The reason behind solving this problems in this method is the basics of long division method, since $x - 3 = 0$ is a root thus it can be considered as a divisor and the cubic polynomial will be a dividend, surely when a degree one polynomial divides a degree 3 polynomial a degree 2 polynomial will be obtained and this degree two polynomial is simply a quadratic equation. The quadratic equation thus obtained can simply be solved using Dharacharya formula as well instead of factorization, in this any quadratic equation of the form $a{x^2} + bx + c = 0$ has roots as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.