Question

Write the sum and product of the zeroes of the polynomial $p\left(x\right)=x^3-4x$.

Answer 1

Step 1. Write the value of $a,b,c\text{and}d$ for the polynomial $p\left(x\right)=x^3-4x$:

Here, the polynomial $p\left(x\right)=x^3-4x$.

We know that, the cubic polynomial in the form of $p\left(x\right)=a{x}^3+b{x}^2+cx+d$

$\Rightarrow p\left(x\right)=x^3+0x^2-4x+0$

On comparing, we get

$\therefore a=1,b=0,c=-4,d=0$

Step 2. Find the sum and product of the zeroes:

$\begin{array}{rcl}\text{⇒sum of zeroes}& =& -\frac{b}{a}\\ & =& -\frac{0}{1}\\ & =& 0\end{array}$

$\begin{array}{rcl}\text{⇒product of zeroes}& =& -\frac{d}{a}\\ & =& -\frac{0}{1}\\ & =& 0\end{array}$

Hence, the sum of the zeroes is 0, and product of the zeroes is 0.