Question

The multiple roots of ${{x}^{5}}-3{{x}^{4}}-5{{x}^{3}}+27{{x}^{2}}-32x+12=0$
(a)1,2
(b)2,3
(c)3,4
(d)4,1

Answer 1

Hint: From the question given we have to find the multiple roots of the ${{x}^{5}}-3{{x}^{4}}-5{{x}^{3}}+27{{x}^{2}}-32x+12=0$. To solve these types of problems we have to substitute the given options in the given equation, this is the simple technique to solve this type of problems by checking the options given in the question.

Complete step-by-step solution:
From the question given we have to find the multiple roots of
$\Rightarrow {{x}^{5}}-3{{x}^{4}}-5{{x}^{3}}+27{{x}^{2}}-32x+12=0$
Let this equation be $f\left( x \right)$, that is
 $\Rightarrow f\left( x \right)={{x}^{5}}-3{{x}^{4}}-5{{x}^{3}}+27{{x}^{2}}-32x+12=0$
To solve these types of problems we have to substitute the given options in the given equation, this is the simple technique to solve this type of problems by checking the options given in the question.
First, we will check with the first option
(a)1,2
Now, we will substitute the 1 in the equation $f\left( x \right)$
$\Rightarrow {{x}^{5}}-3{{x}^{4}}-5{{x}^{3}}+27{{x}^{2}}-32x+12=0$
 By substituting 1 we will get,
$\Rightarrow {{1}^{5}}-3\times {{1}^{4}}-5\times {{1}^{3}}+27\times {{1}^{2}}-32\times 1+12$
$\Rightarrow 1-3-5+27-32+12$
By further simplification we will get,
$\Rightarrow 40-40$
$\Rightarrow 0$
Therefore $f\left( 1 \right)=0$
Now we will substitute the 2 in the equation $f\left( x \right)$
By substituting 2 we will get,
$\Rightarrow {{2}^{5}}-3\times {{2}^{4}}-5\times {{2}^{3}}+27\times {{2}^{2}}-32\times 2+12$
$\Rightarrow 32-48-40+108-64+12$
$\Rightarrow 152-152$
$\Rightarrow 0$
Therefore, $f\left( 2 \right)=0$
Now we will check the numbers in remaining options they are 3 and 4
Now we will substitute 3 in equation,
$\Rightarrow f\left( 3 \right)={{3}^{5}}-3\times {{3}^{4}}-5\times {{3}^{3}}+27\times {{3}^{2}}-32\times 3+12$
$\Rightarrow f\left( 3 \right)=24\ne 0$
Now we will substitute 4 in equation,
$\Rightarrow f\left( 4 \right)={{4}^{5}}-3\times {{4}^{4}}-5\times {{4}^{3}}+27\times {{4}^{2}}-32\times 4+12$
$\Rightarrow f\left( 4 \right)=252\ne 0$
Therefore, by this we can say that the roots of the given equation is 1, 2 that is $f\left( 1 \right)=0$ and $f\left( 2 \right)=0$
Option (a) is correct.

Note: Students should know the factorization of polynomials, and to find roots of the given polynomial in the above question it is difficult to factorize so the option checking is the best method to solve questions like this.