Chapter 2, Problem 3CT

To determine

Calculate all the zeros of the function and also find the multiplicity of each zero.

Answer to Problem 3CT

The all the zeroes of the functions are $x=0andx=\frac{3}{2}$ and the Multiplicity $1$ is $x=0$ and Multiplicity $2$ is $x=\frac{3}{2}$ .

Explanation of Solution

Given:

It is given in the question that the function is $f(x)=4x^3-12x^2+9x$ .

Concept Used:

In this, use the concept of only cubic polynomial and basic factorization.

Calculation:

Here, the function is $f(x)=4x^3-12x^2+9x$ .

Now, factor the above function:

  $\begin{array}{}$ f(x)=4x^3-12x^2+9x$$ f(x)=x(4x^2-12x+9)$$ f(x)=x{(2x-3)}^2$\end{array}$

Now, to find zero , make $f(x)=0$ :

  $\begin{array}{l}f(x)=0\\ $ x{(2x-3)}^2=0$$ x=0;{(2x-3)}^2=0$x=0;x=\frac{3}{2}\end{array}$

Now, the Multiplicity $1$ is $x=0$ and Multiplicity $2$ is $x=\frac{3}{2}$ .

Conclusion:

The all the zeroes are $x=0andx=\frac{3}{2}$ and the Multiplicity $1$ is $x=0$ and Multiplicity $2$ is $x=\frac{3}{2}$ .