Question

Locating zeros of polynomial functions

Answer 1

Answer:

$a.\:\:-1,1$

Step-by-step explanation:

The given polynomial function is;

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

According to the rational roots theorem, the possible rational roots are

$\pm \frac{1}{2} ,\pm1,\pm2$

We now use the remainder theorem to obtain;

$f(1)=2(1)^4-(1)^3+(1)-2$

$f(1)=2-1+1-2$

$f(1)=0$

This implies that, $x=-1$ is a zero.

Also,

$f(-1)=2(-1)^4-(-1)^3+(-1)-2$

$f(-1)=2+1-1-2$

$f(-1)=0$

This means that, $x=1$ is also a zero.

$f(\frac{1}{2})=-\frac{7}{4}$

$f(-\frac{1}{2})=-\frac{7}{4}$

$f(2)=2$

$f(-2)=2$

Hence the zeros are $-1,1$