Question

The multiplicity of the root x = 1 for the function $f(x)={{x}^{2}}{{(x+1)}^{3}}{{(x-2)}^{2}}(x-1)$ is
$\begin{align}
  & \text{a) 1} \\
 & \text{b) 2} \\
 & \text{c) 3} \\
 & \text{d) 4} \\
\end{align}$

Answer 1

Hint: The multiplicity of roots is the number of times the roots occur in the complete factorization of the polynomial. Hence to find the multiplicity of root x = 1, we will check how many times the root occurs in polynomial.

Complete step by step answer:
Now we are given with the equation $f(x)={{x}^{2}}{{(x+1)}^{3}}{{(x-2)}^{2}}(x-1)$
Now the multiplicity of roots is the number of times the roots occur in the complete factorization of the polynomial.
Now we can write this polynomial as $f(x)=x(x)(x+1)(x+1)(x+1)(x-2)(x-2)(x-1)$
Hence for this polynomial the roots are 0, -1 2, 1
Since at x = 0, -1, 2 and 1 we get the value of f(x) equal to 0
Now let us check the multiplicity
Now for the root x = 0.
In the polynomial x occurs 2 times hence multiplicity of x = 0 is 2
Now for the root x = -1.
In the polynomial (x + 1) occurs 3 times hence multiplicity of x = -1 is 3.
Now for the root x = 2.
In the polynomial (x – 2) occurs 2 times hence multiplicity of x = 2 is 2
Now for the root x = 1.
In the polynomial (x – 1) occurs 1 times hence multiplicity of x = 1 is 1
Hence the multiplicity of x = 1 is 1.

Note:
Now the roots with multiplicity 1 are called simple roots hence 1 is a simple root. Also we can directly answer the question by just checking the power of the required root. For example if ${{(x-2)}^{3}}$ occurs in the equation then the multiplicity of 2 is 3.