Question
Show that the polynomial $f(x)=x^4+4 x^2+6$ has no zeroes.
✓
Answer
Let $t = x^2$
So, $f(t) = t^4 + 4t^2 + 6$
Now, to find the zeros, we will equate f(t) = 0
$\Rightarrow t^4 + 4t^2 + 6 = 0$
Now, $\text{t}=\frac{-4\pm\sqrt{16-24}}{2}$
$=\frac{-4\pm\sqrt{-8}}{2}$
$=2\pm\sqrt{-2}$
i.e., $\text{x}^2=-2\pm\sqrt{-2}$
$\Rightarrow\text{x}=\sqrt{-2\pm\sqrt{-2}},$ which is not a real number.
The zeros of a polynomial should be real number s.
$\therefore$ The given f(x) has no zeros.
So, $f(t) = t^4 + 4t^2 + 6$
Now, to find the zeros, we will equate f(t) = 0
$\Rightarrow t^4 + 4t^2 + 6 = 0$
Now, $\text{t}=\frac{-4\pm\sqrt{16-24}}{2}$
$=\frac{-4\pm\sqrt{-8}}{2}$
$=2\pm\sqrt{-2}$
i.e., $\text{x}^2=-2\pm\sqrt{-2}$
$\Rightarrow\text{x}=\sqrt{-2\pm\sqrt{-2}},$ which is not a real number.
The zeros of a polynomial should be real number s.
$\therefore$ The given f(x) has no zeros.
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