91Ó°ÊÓ

The given function is$f\left(x\right)=2x^4+7x^3+x^2-7x-3$

Since the degree of the function is 4 so it has atmost 4 real zeroes.

By descarte's rule of signs, there is only 1 positive zeroe as there is only 1 variation in sign of the function.

$f(-x)=2x^4-7x^3+x^2+7x-3$

So, there is 3 or 1 negative zero as there are 3 variations in sign.

As the leading coefficient of the function, $a_4=2$and the constant term is $-3$, the potential rational zeroes are $Â\pm 1,Â\pm 3,Â\pm \frac{1}{2}andÂ\pm \frac{3}{2}$

Test the potential zero 1, using the synthetic division.

Now, the polynomial $2x^3+9x^2+10x+3$is again factored using synthetic division by taking the value $-1$

The remainder of this is 0. So, $-1$is a zero and $x+1$is a factor of the function.

Using the above entries in the bottom row of the synthetic division, the function can be factored.

$f\left(x\right)=(x-1)(x+1)(2x^2+7x+3)$

The second degree polynomial can also be afctored and the factors are $(2x+1)(x+3)$

Use the zero product property to find the value of $x$

$$\begin{gathered} 2x+1=0 \\ 2x=-1 \\ x=-\frac{1}{2} \\ and \\ x+3=0 \\ x=-3 \end{gathered}$$

Thus, the real zeroes are$\left\{-3,-1,-\frac{1}{2},1\right\}$