# Find all the real and complex zeros of the following polynomials.f(x)=x^4-4x^3+7^2-16x +12

Find all the real and complex zeros of the following polynomials.
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{4}{x}^{{3}}+{7}^{{2}}-{16}{x}+{12}$$

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Neelam Wainwright

It is given that, $$\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{4}{x}^{{3}}+{7}^{{2}}-{16}{x}+{12}$$
We have to find all the real and complex zeros of the polynomial.
We have, $$\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{4}{x}^{{3}}+{7}{x}^{{2}}-{16}{x}+{12}$$
For the zeros of polynomial,
$$\displaystyle\rightarrow{x}^{{4}}-{4}{x}^{{3}}+{7}{x}^{{2}}-{16}{x}+{12}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{3}\right)}{\left({x}^{{3}}-{x}^{{2}}+{4}{x}-{4}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{3}\right)}{\left({x}-{1}\right)}{\left({x}^{{2}}+{4}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{3}\right)}={0}$$ or $$\displaystyle{\left({x}-{1}\right)}={0}$$ or $$\displaystyle{\left({x}^{{2}}+{4}\right)}={0}$$
$$\displaystyle\Rightarrow{x}={3}$$ or $$\displaystyle{x}={1}$$ or $$\displaystyle{x}^{{2}}=-{4}$$
$$\displaystyle\Rightarrow{x}={3}$$ or $$\displaystyle{x}={1}$$ or $$\displaystyle{x}=\pm{2}{i}$$
There are four zeros of the polynomial
Two real numbern $$x=1,3$$
Two complex number $$\displaystyle={2}{i},-{2}{i}$$